Q: Showing that $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$ implies $f$ is constant. Let $E \subset \mathbb{R}^2$ be a connected convex open subset with $(0, 0) \in E$, and $f(x, y)$ be a differentiable function on $E$. Suppose $f$ satisfies $$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$$ then prove $f$ is constant on $E$.

We are given $E$ is convex and connected with $(0,0) \in E$, so it only seems natural to start with a line from the origin to $(x, y)$. Let this be $\gamma(t) = (tx, ty)$. We wish to show that the derivative across this line is zero so that $f(0,0) = f(x,y)$. So, take $h(t) = f(\gamma(t))$ and $h'(t) = f'(\gamma (t))\gamma'(t)$.
Now
$f'(\gamma (t)) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} = x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$
$\gamma'(t) = \gamma$ since $\gamma$ is a linear operator.
Now do I just say that the whole thing equals to zero and be done with it? 

A: You shoud be more careful with your notation, which is wrong at some points. In particular, $f'(\gamma(t))$ should be a vector and not a number.
Another thing: your differential equation reads
$$x \, \frac{\partial f}{\partial x}(x,y) + y \, \frac{\partial f}{\partial y}(x,y) = 0.$$
In particular, there are the same $x$ and $y$ in front of the derivatives and as arguments, where the derivative is evaluated. This is wrongly used in your proof.
