Gradient vector parallel to postition vector Let $f:\mathbb{R}^3\to \mathbb{R}$ be a differentiable fuction such that $\nabla f(x,y,z)$ is parallel to $(x,y,z)$ for all $(x,y,z)\in \mathbb{R}^3$. Show that, $f(a,0,0)=f(-a,0,0)$ for all $a\in \mathbb{R}^{+}$.
My attempt:
I know that $\nabla f(x,y,z)$ is parallel to $(x,y,z)$ if and only if $\nabla f(x,y,z)\times (x,y,z)=0$. Usign that fact, I found the follwing system of equations,
$$\begin{cases}
zf_y=yf_z,\\
zf_x=xf_z,\\
yf_x=xf_y.
\end{cases}$$
How can I prove the proposition from this system? Is this the correct approach? Thanks!
 A: In fact something much stronger is true, given your assumptions: the value of $f$ depends only on the distance from the origin. Picture a sphere centered on the origin. At any point on the sphere, the gradient of $f$ points straight out (or in), in particular it has no component tangential to the sphere. Therefore, at that point, the value of $f$ does not change as you move along the surface of the sphere. Since this is true all over the sphere, the value of $f$ is the same all over the sphere.
You could turn this into a proof by for example taking two points on the sphere, and drawing a curve $\gamma$ between them lying entirely on the sphere. You could then show that $\frac d{dt} f(\gamma(t))=0$ for all $t$.
I'm not sure if there's a way of proving your statement which doesn't involve more or less proving this stronger statement. You cannot just consider the restriction of $f$ to the line $t\to(t, 0, 0)$, since at that point you've thrown out all the important information and are basically trying to prove that any single-variable function is even. You're going to have to leave that line somehow and prove that $(a, 0, 0)$ and $(-a, 0, 0)$ have the same value in some way that involves the behavior of $f$ off of the line, for example by drawing a path between them.
