I have the following problem and I would like to know if my answer is correct:
Let $\Omega \subset \mathbb{R}^n$ be an open set and $f:\Omega \rightarrow \mathbb{R}, f \in C^2(\Omega)$. Let $a \in \Omega$ be a critical point of $f$ such that the determinant of the Hessian matrix is non zero. Then there exists $U$ open set such that $a \in U$ and $a$ is the only critical point of $f$ in $U$.
Proof:
Call the differential of $f$, $Df:\mathbb{R}^n \rightarrow L(\mathbb{R}^n, \mathbb{R})$, $g$. As $f$ is $C^2(\Omega)$, $g$ is differentiable in $\Omega$, indeed $g \in C^1(\Omega)$. We also have that $g(a)=0$ and $Dg(a)$ is invertible. Now we use the inverse function theorem to $g$. We therefore get an open set $U$ such that $a \in U$ and $g$ restricted to $U$ is biyective, in particular inyective. From that we get that no other point $x \in U$ verifies $g(x)=0$, and therefore there are no critical points of $f$ in $U$ appart from $a$.