F(x,y,z) Symmetry in Scalar Function g: Counterexample or Confirmation?

[hauho195]

Today I've tried to investigate properties of a function $f(x_1,x_2,x_3)$ satisfying $\nabla\times(f\mathbf{x})=0$ in $\mathbb{R}^3$, where some degrees of the differentiability are assumed if needed.

By some basic procedures, I've deduced that: there is a scalar function g such that $f=\frac{1}{x_i}\frac{\partial{g}}{\partial{x_i}}$ for i=1,2,3. Then I guess f is symmetric(i.e., f(x,y,z)=f(x,z,y)=f(y,x,z)=f(y,z,x)=f(z,x,y)=f(z,y,x)). Is it true? If not, is there a counterexample?

 

[mfb]

This is right, and you can even show that f can depend on $\sqrt{x_1^2+x_2^2+x_3^2}$ only.