Let $f,g,h$ be three real-valued functions on an open subset of $\mathbb R^n$ such that $$f = gh.$$
Suppose that $f$ and $g$ are smooth, i.e., infinitely differentiable. Under what conditions does this imply that $h$ is smooth?
The following counterexample $$f(x) = x,\qquad g(x) = x^2,\qquad h(x)=\left\{\begin{matrix}1/x &x\neq 0\\ 0 &x=0\end{matrix}\right.$$ show that this is not necessarily true. On the other hand, $g$ being is nowhere vanishing would be a sufficient condition since in that case we could write $h = f/g$.
Is there anything interesting to say in between?
I don't think there's much of an in between: Already if $g$ has just one root, then $h$ doesn't have to be smooth anymore. For example, consider $g=f=\text{id}_{\mathbb{R^n}}$ and $h=1_{\mathbb{R^n}}-1_{\{0\}}$. Then $f,g$ are smooth and $f=gh$ but $h$ is obviously not even differentiable.