checking the diffeomorphism between 2 surfaces I tried to show that a surface $x^4+y^2+z^2=1$ and the unit sphere are diffeomorphic.

Since a diffeomorphism between them is not chosen easily, I would apply a theorem using invertibility of Jacabian.

For this process, I must know a surface patch of $x^4+y^2+z^2=1$, but I don't know that.

Q1. How do I choose the surface patch?

Q2. Is my plan is correct to show that they are diffeomorphic?
Let $M=\{(x,y,z): x^4+y^2+z^2=1\}$. The (radial projection) map $F\colon \Bbb R^3-\{0\}\to S^2$ given by $F(x) = x/\|x\|$ is smooth, hence restricts to a smooth map $f\colon M\to S^2$. It is easy to check that $f$ is a bijection. You can then prove that it is a diffeomorphism by noting that for any $p\in M$, $$T_pM\cap \{tp: t\in\Bbb R\}=\{0\} \tag{$\star$}$$ (i.e., $T_pM$ is transverse to the line spanned by $p$), and so $\ker df_p = \ker dF_p\cap T_pM = \{0\}$. Thus, by the inverse function theorem, $f$ is a local diffeomorphism, and we're done.

EDIT: I guess I should elaborate a bit on the claim ($\star$). If $p=(a,b,c)$, the normal vector to $T_pM$ is $(2a^3,b,c)$. If the line spanned by $p$ were contained in $T_pM$, we'd have $(2a^3,b,c)\cdot (a,b,c) = 0$, which leads us to the system
\begin{align*}
2a^4+b^2+c^2&amp;=0 \\ a^4+b^2+c^2&amp;=1
\end{align*}
This means $a^4=-1$, which has no solutions.

It is my suggestion (whenever possible) to avoid explicit chart-checking and to work with the implicit representation of the manifold.
It is obvious that the map sending $x$ to $\pm x^2$ defines diffeomorphisms between the upper and the lower hemispheres of your surfaces. You just need to check that things are OK on the equator. Can you? 

A different way is to check that both surfaces are strictly convex, then they are diffeomrophic.