Q: Find all $P_0= (x_0, y_0, z_0)$ on $z = x + y^2$ so that the angle between normal vectors at $P_0$ and $(0,1,0)$ is $\pi/4$. Question:

Find all $P_0= (x_0, y_0, z_0)$ on $z = x + y^2$ so that the angle
  between normal vectors at $P_0$ and $(0,1,0)$ is $\pi/4$.

My attempt:
Let $F(x,y,z) = z - x - y^2$.Then the condition that the angle between the normal vector at $P_0$ and $j$ can be expressed as 
$$\nabla F(P_0) \cdot j = |\nabla F(P_0)| 1/\sqrt{2},$$
hence
$$-2y_0 = \sqrt{1^2 + 4y_0^2 + 1^2} * 1/\sqrt{2},\quad (2)$$
but this gives that $y_0^2 = -1/2$.
Therefore, my first question is that where is the mistake ?
Secondly, if the LHS of equation $(2)$ had contained other terms ($x_0$, and $z_0$), would subtracting LHF from RHS and defined that difference as $G(x,y,z)$ and using Lagrange multiplier method work ?

A: Question:
Find all $P_0=(x_0,y_0,z_0)$ on $z=x+y^2$ so that the angle between normal vectors at $P_0$ and $(0,1,0)$ is $\pi/4$.
Answer:
In response to the first questioner I think the question is requiring you to find the subset of the surface where the normals are at $\pi/4$ to $(0,1,0)$ since $(0,1,0)$ is not on the surface.
In response to the second questioner the surface is insufficiently specified to answer this question uniquely. There are two possible and distinct answers depending on the orientation of the surface.
The surface we shall consider is specified by the points $\{(x,y,z)\in\mathbb{R}^3|F(x,y,z)=z-x-y^2=-z+x+y^2=0\}$.
Case 1:
The first possible orientation we shall consider fixes the normal to this surface at any point $\vec{P_0}=(x_0,y_0,z_0)$ to be a positive multiple of $\vec{\nabla}{F}(\vec{P_0})=(-1,-2y_0,1)$. For such a normal to be at angle $\pi/4$ to the point $(0,1,0)$ 
$$
\vec{\nabla}{F}(\vec{P_0}).(0,1,0)=||\vec{\nabla}{F}(\vec{P_0})|| \ ||(0,1,0)|| 
\cos{(\pi/4)}.
$$
Requiring this renders the equation
$$
-2y_0=\sqrt{1+2y_0^2}.
$$
This may only be solved when $y_0$ is negative given that the surface is real. Squaring both sides and rearranging we arrive at the condition 
$$
2y_0^2=1,
$$
which is only consistent with previous steps when $y_0=-1/\sqrt{2}$. Hence the only possible points on the surface where the normals are at $\pi/4$ to $(0,1,0)$ are elements of $\{(x,-1/\sqrt{2},x+1/2)|x\in\mathbb{R}\}$ that is the curve on the surface with points $(x,y,z)$ such that $y=-1/\sqrt{2}$ and $z=x+1/2$.
Case 2:
The second possible orientation we shall consider fixes the normal to this surface at any point $\vec{P_0}=(x_0,y_0,z_0)$ to be a positive multiple of $\vec{\nabla}{F}(\vec{P_0})=(1,2y_0,-1)$. For such a normal to be at angle $\pi/4$ to the point $(0,1,0)$ 
$$
\vec{\nabla}{F}(\vec{P_0}).(0,1,0)=||\vec{\nabla}{F}(\vec{P_0})|| \ ||(0,1,0)|| 
\cos{(\pi/4)}.
$$
Requiring this renders the equation
$$
2y_0=\sqrt{1+2y_0^2}.
$$
This may only be solved when $y_0$ is positive given that the surface is real. Squaring both sides and rearranging we arrive at the condition 
$$
2y_0^2=1,
$$
which is only consistent with previous steps when $y_0=1/\sqrt{2}$. Hence the only possible points on the surface where the normals are at $\pi/4$ to $(0,1,0)$ are elements of $\{(x,1/\sqrt{2},x+1/2)|x\in\mathbb{R}\}$ that is the curve on the surface with points $(x,y,z)$ such that $y=1/\sqrt{2}$ and $z=x+1/2$.
