Let $f:\mathbb{R}^n\to\mathbb{R}$ be a smooth function and let $x_0\in \mathbb{R}^{n+1}$ be a critical point of $f$ such that $f(x_0)$ is a local maximum.
Question: Under which conditions does it follow that the Hessian $\mathrm{Hess}_f(x_0)$ has nonzero determinant? Note that this is clearly false if $f$ is a constant function, since then $\mathrm{Hess}_f$ is identically zero. If we assume $f$ is nonconstant, is this sufficient to ensure that $\det \mathrm{Hess}_f(x_0)\neq 0$?
The second derivative test says that $\mathrm{Hess}_f(x_0)$ being positive definite is sufficient for $x_0$ to be a maximum. So the question is in some sense a converse to that theorem, although I am only interested $\det \mathrm{Hess}_f(x_0)$ being nonzero.
If if $f$ is not constant, the determinant of the Hessian may be $0$. Take, for instance, $f(x,y)=-x^4-y^4$. Then $f$ has an absolute maximum at $(0,0)$ and, even so, the Hessian of $f$ at $(0,0)$ is the null matrix.