$f(x,y,z)$ must be equal at $(0,0,a)$ and $(0,0,-a)$ when $\nabla f(x,y,z)$ is always parallel to $(x,y,z)$

Question

This vector calculus exercise from Apostol's book has taken me some hours without any advance. Can you give some hints please?

If $\nabla f(x,y,z)$ is always parallel to $(x,y,z),$ prove that $f$ must take equal values at the points $(0,0,a)$ and $(0,0,-a).$

The exercise appears in the section which develops chain rule and the proof that the gradient vector is normal to the level surfaces.

Answer 1

Prove that $f$ is constant on the spheres centered in $(0,0,0)$. In fact if you take any curve $\gamma(t)$ such that $|\gamma(t)|=R$ you have $\langle \gamma'(t),\nabla f(\gamma(t))\rangle = 0$ since $\gamma'$ is tangent to the sphere while $\nabla f$ is orthogonal.