Show by definition that $M=\{(x,y,z)|36x^2+4y^2-9z^2=36\}$ is a surface in $\Bbb R^3$.
Definition
A surface in $\Bbb R^3$ is a subset $M$ of $R^3$ such that for each point $p$ of $M$ there exists a proper patch (a 1-1 regular mapping of an open set $D$ of $\Bbb R^2$ into $\Bbb R^3$ for which the inverse function is continuous) in $M$ whose image contains a neighborhood of $p$ in $M$.
I came up with the patch $x(u,v)=(\operatorname{cosh}u\operatorname{cos}v,3\operatorname{cosh}u\operatorname{sin}v,2\operatorname{sinh}u)$ on $\Bbb R^2$. Then this is a regular mapping, however, it is not 1-1,and if I try to make it 1-1 by restricting the domain that it does not cover the whole surface, nor can I prove that the inverse is continuous.
How can I prove that $M$ is a surface directly from definition by finding proper patches in this case? I would greatly appreciate it if anyone could help me.
The patch you found is good. Let's write everything neatly. We have:
$${\bf x} \colon \Bbb R \times \left]0,2\pi\right[ \to M \\ {\bf x}(u,v) = (\cosh u \cos v, 3\cosh u \sin v, 2\sinh u)$$
This won't cover the whole surface: one meridian will be left out: the curve $(\cosh u,0,2\sinh u)$, which would correspond to ${\bf x}(u,0)$, but our domains must be open sets. You can take $\epsilon > 0$ and use the same expression to define a patch with domain $\Bbb R \times \left]-\epsilon,\epsilon\right[$ to cover what is left. Everything will work. And the transition map will be the identity.
To prove that ${\bf x}^{-1}$ is continuous, it suffices to check that $\bf x$ is bijective, differentiable, and ${\rm d}{\bf x}_{(u,v)}$ has maximum rank for all $(u,v)$ in the domain. In other words, check that: $$\frac{\partial {\bf x}}{\partial u}\times \frac{\partial {\bf x}}{\partial v}(u,v) \neq {\bf 0}.$$