Let $f: V \to \mathbb{R}$ of class $C^{2}$ and $b \in V$ a critical point of $f$. If $\varphi: U \to V$ is a diffeomorphism $C^{2}$ with $\varphi(a) = b$, then the hessian of $f$ in the point $b$ and the hessian of $f \circ \varphi$ in the point $a$ has a same rank.
I wanted a way to go. I tried to do some things, but I didn't have a good ideas. I know I need to use Inverse Function Theorem, but I don't know how to use it. I don't want a solution for the question, I just want an idea to continue. Thanks for the advance!
EDIT (because of Daniele's comment). I have to use the Inverse Function Theorem and this implies that $\varphi$ is a local diffeomorphism iff $\varphi'(x): U \to V$ is an isomorphism $\forall x \in U$ so, I can understand that $\nabla\varphi: TU_{x} \to TV_{\varphi(x)}$ is an isomorphism. But I still cannot see how to use this to conclude the question. Is probably a little detail that I cannot see.
I believe, based on what you have, that you just need a fact from linear algebra...
Namely, if $T$ is a linear transformation, and $I$ an isomorphism, then $\operatorname{rank}({T\circ I})=\operatorname{rank}T$.
This follows from the fact that $I(V)=V$ and $\operatorname{rank}T=\dim(\operatorname{image}T)$...
So, let $T$ be your Hessian at a point $a$; and $I$ the isomorphism, $\phi'$,from the tangent space at point $a$ to point $\phi(a)=b$, where $\phi$ is your diffeomorphism...