I have the following set of $R^3$:
$$ S=\{(x,y,z) \in \mathbb{R}^3: \, x^2+y^2-z^3=1\} $$
It is immediate to see that $S$ is a regular surface. How may I find an atlas? When $z \neq 0$, as $z=\sqrt[3]{x^2+y^2-1}$, we have the parameterization $x(u,v)=(u,v,\sqrt[3]{u^2+v^2-1})$ but what happens when $z=0$?
Hint: At least one of $x$, $y$ and $z$ is not zero. So the open sets $\{x< 0\}$, $\{x>0\}$, $\{y < 0 \}$, $\{y>0\}$ and $\{z\neq 0\}$ cover $S$.
Since you already found a parametrization for $z\neq 0$, I'm sure you will find parametrization for the coordinate neighborhoods $x < 0$, $y < 0$ and so on. Then you are left to show that the change of coordinates is smooth.