Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 2F (Geometry)

The condition $D{f}_p\ne 0$ says that $0$ is a regular value of $f$. The regular level set theorem therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in ${\mathbb{R}}^3$, hence a smooth surface.

Forthe gradient isOn $F^{-1}(0)$, $x^2+y^2=\cosh (z^2)\ge 1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne 0$ there, and the given set is a smooth surface.

Not every smooth surface in ${\mathbb{R}}^3$ is a global zero set. Every set $f^{-1}(0)$ is closed because $f$ is continuous, whereas the open unit diskis a smooth surface but is not closed in ${\mathbb{R}}^3$.