I'm trying to understand the derivation of Wiener deconvolution given on its Wikipedia page.  In the last couple steps under the derivation section, they take the derivative with respect to $G(f)$ of an equation that has both $G(f)$ and $G^\ast(f)$ in it.  They simply state that $G^\ast (f)$ acts as a constant in the differentiation.  However, it seems to me that if you don't treat $G(f)$ as a constant, then you shouldn't be able to treat $G^\ast (f)$ as a constant because they are directly related.  

I searched around some looking for an explanation.  I found this page, which seems to agree that the complex conjugate can be treated as a constant.  I also found some stuff about the Cauchy-Riemann equations, which seem to be related.  However, I haven't had any classes on complex analysis and don't understand the intuition behind why this can be done.

Why can the complex conjugate of a variable be treated as a constant when differentiating with respect to that variable?

The nomenclature of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ is confusing because it gives the impression that these are really partial derivatives with respect to two independent variables, $z$ and $\bar{z}$. However, it is clear that $z$ and $\bar{z}$ are not independent.

Differentiable Functions and Conformal Maps

A differentiable function on $\mathbb{R}$ locally looks like a linear function, that is, there is a real constant, called $f'(x)$, so that for small $h$,
$$
f(x+h)=f(x)+f'(x)h+o(h)\tag{1}
$$
Analogously, a differentiable function on $\mathbb{C}$ satisfies $(1)$ for some complex number $f'(x)$.

Multiplication on $\mathbb{C}$ acts as a rotation and radial scale when viewed as an action on $\mathbb{R}^2$. Thus, if $f$ is differentiable on $\mathbb{C}$,
$$
f(z+h)-f(z)=f'(z)h+o(h)\tag{2}
$$
That is, when $h$ is small, $h\mapsto f(z+h)-f(z)$ looks like a scaled rotation. For this reason, a differentiable function on $\mathbb{C}$ is called conformal: small features are replicated (scaled and rotated) and angles are preserved.

Complex Conjugation and Orientation Reversal

Complex conjugation, $z\mapsto\bar{z}$, is an orientation reversing isometry. Thus, when composed with a conformal map, either before or after, the composition is an orientation-reversing conformal map. Furthermore, double composition yields an orientation-preserving conformal map; for example, if $f(z)$ is conformal, then so is $\overline{f(\bar{z})}$.

As a function on $\mathbb{R}^2$, complex conjugation can be represented by the matrix $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.

Conformal and Conjugate Conformal

The partial derivatives of a general differentiable function on $\mathbb{R}^2$ given by $x+iy\mapsto u+iv$ are usually given in a $2\times2$ Jacobian matrix:
$$
\frac{\partial(u,v)}{\partial(x,y)}=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}\\\frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}\end{bmatrix}\tag{3}
$$
The Cauchy-Riemann equations specify that $\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$, which agrees with the following basis for the orientation-preserving conformal Jacobians on $\mathbb{R}^2$:
$$
\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}\tag{4}
$$
Note that the determinant of any linear combination of these matrices has positive determinant (thus orientation is preserved).

The following basis for the orientation-reversing conformal Jacobians on $\mathbb{R}^2$ follows by composing conjugation with $(4)$:
$$
\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}\tag{5}
$$
Note that the determinant of any linear combination of these matrices has negative determinant (thus orientation is reversed).

Using $(4)$ and $(5)$, we can break any Jacobian into conformal and conjugate conformal parts. Using the component-wise orthogonality that exists among the bases, we can write the conformal part as
$$
\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{6}
$$
and the conjugate conformal part as
$$
\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&-1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\1&0\end{bmatrix}\tag{7}
$$
$\dfrac{\partial}{\partial z}$, $\dfrac{\partial}{\partial\bar{z}}$, and Quaternions

The definitions of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ say
$$
\begin{align}
\frac{\partial}{\partial z}(u+iv)
&=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)(u+iv)\\
&=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{8}
\end{align}
$$
and
$$
\begin{align}
\frac{\partial}{\partial\bar{z}}(u+iv)
&=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)(u+iv)\\
&=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{9}
\end{align}
$$
The space of $2\times2$ Jacobians has $4$ dimensions, so trying to represent these $4$ dimensions with the $2$ dimensions of $\mathbb{C}$, using $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$, obscures something.

There is a common matrix representation of the complex numbers as $2\times2$ real matrices where
$$
\begin{align}
\mathbf{1}&\leftrightarrow\begin{bmatrix}1&0\\0&1\end{bmatrix}\tag{10}\\
\mathbf{i}&\leftrightarrow\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{11}
\end{align}
$$
However, there is also a matrix representation of the quaternions as $2\times2$ complex matrices where, in addition to $(10)$ and $(11)$,
$$
\begin{align}
\mathbf{j}&\leftrightarrow\begin{bmatrix}i&0\\0&-i\end{bmatrix}\tag{12}\\
\mathbf{k}&\leftrightarrow\begin{bmatrix}0&-i\\-i&0\end{bmatrix}\tag{13}
\end{align}
$$
Embed $(8)$ and $(9)$ in the quaternions to get
$$
\left(\frac{\partial}{\partial z}(u+iv)\right)\mathbf{1}
=\frac{\mathbf{1}}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{\mathbf{i}}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{14}
$$
and
$$
\left(\frac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j}
=\frac{\mathbf{j}}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{\mathbf{k}}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{15}
$$
Finally, substituting $(10)$-$(13)$ into $(14)$ and $(15)$, it becomes apparent, upon comparison with $(6)$ and $(7)$, that $\left(\dfrac{\partial}{\partial z}(u+iv)\right)\mathbf{1}$ represents the conformal part of the Jacobian and $\left(\dfrac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j}$ represents the conjugate conformal part.

Conclusion

For a general $f:\mathbb{C}\mapsto\mathbb{C}$, $\dfrac{\partial}{\partial z}f$ can be mapped to the conformal part of the $2\times2$ Jacobian, $\dfrac{\partial f}{\partial z}=\dfrac{\partial(u,v)}{\partial(x,y)}$, and $\dfrac{\partial}{\partial\bar{z}}f$ can be mapped to the conjugate conformal part. It is merely convenience of notation that we write $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ because $\dfrac{\partial}{\partial z}f\;\mathrm{d}z+\dfrac{\partial}{\partial\bar{z}}f\;\mathrm{d}\bar{z}=\mathrm{d}f$. However, they are not true partial derivatives, but $2$ pieces of a $2\times2$ Jacobian composed of $4$ partial derivatives.

So, to answer the question asked, $z\mapsto\bar{z}$ is conjugate conformal, so $\frac{\partial}{\partial z}\bar{z}=0$; therefore, $\bar{z}$ acts like a constant under $\frac{\partial}{\partial z}$.
