Consider the function $$f: \mathbb{R}^3 \to \mathbb{R} , \ f (x,y,z) = \cos^{2} {(x)} \cdot \sinh ^{2} {(y)} -z^{2} +c$$
For which values of $c$ is the zero-set of $f$ a regular surface?
By definition, a subset $M \subset \mathbb{R}^n$ is called a regular surface if for each point $p \in M$, there exists a neighbourhood $V$ of $p$ in $\mathbb{R}^n$ and a map $x: U \to \mathbb{R}^n$ of an open set $U \subset \mathbb{R}^2$ onto $V \cap M$, such that
$x$ is differentiable;
$x: U \to V \cap M$ is a homeomorphism;
Each map $x: U \to M$ is a regular patch.
I set $f (x,y,z) = \cos^{2} (x) \sinh^{2} (y) -z^{2} +c$ , then $\cos^{2} (x) \cdot \sinh^{2}(y) -z^{2} = -c$.
Also, set $\nabla f = \left( -2 \sin (x) \cos(x) \sinh ^{2} (y) \ 2\cos^{2}(x) \ \sinh (y) \cosh(y) \ -2z \right) = 0 $ to get the critical points $(x,0,0)$, where $x$ can be any real number.
But what's next? How can I apply the definition to such problems? Thanks!
In lay terms, regularity means 1) we don't want edges; 2) we don't want creases; 3) we don't want two parts of surfaces to be glued in points or along lines (no self intersections).
The given surfaces bifurcates at point $c=0$, since when $c\le 0$, $z$ can be anything including $0$; when $c>0$, possible $z$ values are disjoint ($z\le-\sqrt c \cap z\ge\sqrt c$). That gives as 3 cases to consider:
As was said, $z$ values are disjoint (and symmetric), so we only need to consider one part of the surface $z>0$: $$ z = \sqrt{c+\cos^2x \sinh^2 y} $$ That function is defined for all $x,y$ (no edges); cannot touch the other surface (no glued surfaces); and we need to get sure about creases. We take the parametrization of patch $x,y$ and find Jacobian: $$ J=\begin{pmatrix}1&0\\0&1\\z'_x&z'_y\end{pmatrix} $$ Since derivatives can be defined for all $x,y$ the rank of Jacobian is 2 and every patch is regular (no creases).
You can show that your surfaces doesn't intersect planes $y=0$ and $x=\pi/2+\pi n$, so we can consider only part where $y>0$, $-\pi/2 < x < \pi/2$ (other parts are symmetric): $$ y=\sinh^{-1}\sqrt{\frac{z^2-c}{\cos^2x}} $$ Using the same arguments we show that there are no edges, creases and touching surfaces.
Consider a neighbourhood of $(0,0,0)$. There is a surface $z=|\cos x\sinh y|$ and surface $z=-|\cos x\sinh y|$ and they are glued at the point $(0,0)$