I'm trying to find a differentiable function function $g:U \to \mathbb R$ (where $U = \mathbb R^3 - {(0,0,0)}$) such that:
$dg/dx = x(f \circ r)$
$dg/dy = y(f \circ r)$
$dg/dz = z(f \circ r)$
where $f : \mathbb R^+ \to \mathbb R$ is a differentiable function and $r(x,y,z) = \sqrt{x^2+y^2+z^2}$ ($r:U\to\mathbb R$).
Thanks!
$$\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}h(r)={x\over \sqrt{x^2+y^2+z^2}}h(r)+xh'(r)=xf(r)$$ $$h'(r)+{1\over r}h(r)=f(r)\implies (rh(r))'=rf(r)\implies h(r)=\frac{\int{rf(r)\,dr}}{r}$$ So you can let $$g(x,y,z)=\int{rf(r)\,dr}$$ Then $$\frac{\partial g}{\partial x}=rf(r)\frac{\partial r}{\partial x}=rf(r)\frac{x}{r}=xf(r)$$ and similarly with the other two variables.