We are given the function $f(x,y,z):=x^3+y^2-z^2-1$ and have to consider the solution set (a "surface") $S:=\{(x,y,z)\in{\mathbb R}^3\ |\ f(x,y,z)=0\}\ .$ As $\nabla f(x,y,z)=(3x^2,2y,-2z)$ is $\ =(0,0,0)$ only at the origin $O\notin S$, by the implicit function theorem the set $S$ is a smooth surface in the neighborhood of all of its points. Here is a picture of $S$:

For given $y$ and $z$ the equation $f(x,y,z)=0$ has exactly one solution $x=\phi(y,z)\in{\mathbb R}$ which is commonly written as $\phi(y,z)=\root 3\of {1-y^2+z^2}$. Unfortunately along the hyperbola $y^2-z^2=1$ the function $\phi$ is not differentiable as a function of $y$ and $z$.
If we are allowed to use more than one patch to cover all of $S$ we could use three patches as follows: $(x,t)\mapsto\bigl(x,-\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty $(x,t)\mapsto\bigl(x,\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty $(y,z)\mapsto\bigl(\root 3\of{1-y^2+z^2}, y, z\bigr)\qquad\bigl(-\infty