Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function such that $f(0) = 0$ and $\nabla f(0) = O$. It is possible to factorize $f$ as $f(x) = g(x) h(x)$ where $g,h :\mathbb{R}^n \rightarrow \mathbb{R}$ are smooth and $g(0)=h(0)=0$?
I did not manage to conclude anything neither in the case when $f\geq 0$, where seems natural to define $g = h = \sqrt[]{f}$ but as shown in https://mathoverflow.net/questions/105438/square-root-of-a-positive-c-infty-function $\sqrt{f}$ can be non smooth.
Write out the def. of derivative at $x=0 $, for any direction $\ \vec e$: $$\frac{f(0+h\vec e) - f(o) - [0](\vec e)}{h} \longrightarrow 0$$ But the LHS equals $$\frac{f(h\vec e)}{h}:=g(h\vec e) $$ $$g(0):=0$$ Now, we do have that $$\forall h, f(h\vec e)= h.g(h\vec e)$$
$h$ seen as function is already nice: smooth, and zero at zero. So, if you prove that $g$ defined above is smooth at $0$ then you're done (nonzero points follow from smoothness of $f$.)
I think if you take derivatives, you can prove that all exist at 0, which finishes this construction.