Let $S$ be a regular surface $S$ given by the graph of a differentiable function $f:\mathbb{R^2} \rightarrow \mathbb{R}$ and $(\mathbb{R^2}, \varphi)$ an atlas of $S$ such that $\varphi(u,v)=(u,v,f(u,v))$. Consider $(\mathbb{R^2}, \psi)$ where $\psi(x_1,x_2)=(sinh(x_1),x_2^3,f(sinh(x_1),x_2^3))$. Is $(\mathbb{R^2}, \psi)$ an atlas of $S$?
I think it is, so I have tried to prove that $\psi= \varphi \circ h$ Where $h:\mathbb{R^2} \rightarrow \mathbb{R^2}$ is a diffeomorphism, which would prove that $(\mathbb{R^2}, \psi)$ is an atlas.
If we define $h:\mathbb{R^2} \rightarrow \mathbb{R^2}, h(x,y)=(sinh(x),y^3)$ then $\psi= \varphi \circ h$. I've tried to use the inverse function theorem to prove that $h$ is a diffeomorphism:
$h \in C^{\infty}(\mathbb{R^2},\mathbb{R})$ so if $Jac(dh(p)) \neq 0 \ \forall p \in \mathbb{R^2}$ then its inverse is $C^{\infty}(\mathbb{R^2},\mathbb{R})$ and we are done.
$Jac(df(p))=det\begin{bmatrix}cosh(p_1) & 0\\0 & 3p_2^2\end{bmatrix}\neq 0 \iff p_2 \neq 0$. We then have that $h$ is a diffeomorphism in $\mathbb{R^2}-\{ (x,y)\in \mathbb{R^2} : y=0\}$, but we need it to be a diffeomorphism in $\mathbb{R^2}$. I don't know how to continue from here.
What you did is not actually an application of the inverse function theorem.
Your calculation of $Jac(dh(p))$ shows that $h : \mathbb{R}^2 \to \mathbb{R}^2$ is not a diffeomorphism, because it equals zero on all points of the form $(p_1,0)$ (there is a typo in your post, $Jac(df(p))$ should be $Jac(dh(p))$).
But in order for $(\mathbb{R}^2,\psi)$ to be an atlas, it is necessary that $h$ be a diffeomorphism.
To see why, suppose that $(\mathbb{R}^2,\psi)$ is an atlas. We know that the projection function $\pi : S \to \mathbb{R}^2$ defined by $\pi(u,v,f(u,v)) \mapsto (u,v)$ is a diffeomorphism from $S$ to $\mathbb{R}^2$. Therefore, the composition $\pi \circ \psi : \mathbb{R}^2 \to \mathbb{R}^2$ is a diffeomorphism. But $\pi \circ \psi = h$.