Does an unbounded gradient implies a non-vanishing hessian determinant

Question

Let $\Phi:\mathbb{R^{n-1}\to R}$ be a vector function. Suppose that there exists a set $S_1\subset \mathbb R^{n-1}$ on which $G=\nabla\Phi$, the gradient of the function is unbounded.

Does that imply necessarily that there exists a set $S_2$ on which the Hessian determinant $H=\det (\frac{\partial^2\Phi}{\partial x_i\partial x_j})$ or at least one of the eigenvalues is non-vanishing?

Answer 1

Suppose $f(x,y) = x^2.$ Then $\nabla f(x,y) = (2x,0)$ is unbounded. But the determinant of the Hessian matrix of $f$ is $0$ everywhere.