I am reading Jet Nestruev's Smooth Manifolds and Observables book and I am struggling with the exercise 2.12. It states that any smooth function $f:\mathbb{R}^2\to\mathbb{R}$ vanishing on coordinate cross $K=\{x=0\}\cup\{y=0\}$ must be of form $f(x,y)=xyg(x,y)$ for some smooth function $g:\mathbb{R}^2\to\mathbb{R}$.
The exercise is states just after Hadamard's lemma, so I guess I should use it or related ideas somehow.
EDIT. I just managed to prove that
$$f(x,y)=x\int_0^1\frac{\partial f}{\partial x}(tx, y)dt \\ f(x,y)=y\int_0^1\frac{\partial f}{\partial y}(x, ty)dt$$
Thus $f(x,y)=xg_1(x,y)=yg_2(x,y)$. However, I now stuck there.
EDIT 2. Ok, I guess I know the answer. Since $f$ vanishes on the cross $K$, $\frac{\partial f}{\partial x}$ vanishes on $x$-axis. Thus we can apply the same argument as previously and obtain that
$$f(x,y)=xy\int_0^1\int_0^1\frac{\partial^2f}{\partial x\partial y}(tx, sy)dsdt$$
By Hadamard's lemma, $f(x,y) = x\phi(x,y)$ for some smooth $\phi$. Since $f(x,0)=0$ for all $x$, we deduce that $\phi(x,0)=0$ for all $x$ and hence $\phi(x,y)=yh(x,y)$ for some smooth $h$. Thus, $f(x,y)=xyh(x,y)$.