Suppose f:R^3\(Ball of radius 1)--->R is smooth and satisfies f(S^2)=0, ie the unit sphere is a level set of f. does it neccessarily follow that f is a spherically symmetric function?
I don't think so. I have an idea for a counterexample but you will have to verify it for me because I'm not 100%.
The idea is that the modulus function $\displaystyle \mathbb{R}^{3} \rightarrow \mathbb{R}; x \mapsto \|x\|$ is smooth, so if we smoothly stretch $\displaystyle \mathbb{R}^{3}$ and apply a modulus function to the stretched space we should still have a smooth function, but not a spherically symmetric one (provided that our stretching doesn't preserve spheres!).
So here's what I propose. Let $\displaystyle s \in S^{2}$ be any point on the unit sphere, and let $\displaystyle s+ts$ for $\displaystyle t\geq 0, t\in\mathbb{R}$ be the ray eminating from the origin through $\displaystyle s.$ Then $\displaystyle \|ts\|$ is the distance along the ray we have travelled from $\displaystyle s.$ Any point can be expressed in the form $\displaystyle s+ts$ and for any such point the function $\displaystyle x\mapsto t\|s\|$ is a smooth, spherically symmetrical function $\displaystyle \mathbb{R}^{3}/ \{ \text{Ball of radius 1} \} \rightarrow \mathbb{R}$. If we dilate that ray by a factor proportional to the altitude angle and consider the new distance from $\displaystyle s,$ while leaving $\displaystyle S^{2}$ unchanged, then that distance would be a smooth function where $\displaystyle S^{2} \rightarrow 0$ but which is not spherically symmetric.
I guess what I'm saying is, for any $\displaystyle x$ in our domain, let $\displaystyle x = s+ts$ for $\displaystyle s\in S^{2},t\in\mathbb{R}.$ Expressing $\displaystyle s$ as $\displaystyle (1,\theta,\phi)$ in spherical coordinates, then the function $\displaystyle x \rightarrow (|\phi|+1)t \|s\| $ has the desired properties but is not spherically symmetric.
This is more or less a sketch but if you can formalise it a bit, and if you can't find any technical problems with it, then hopefully it will work!