Let $M = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$. I want to show that the function $f: M \to \mathbb{R}^2$ defined by $f(x, y, z) \to (x, y+ z)$ is a local diffeomorphism at the point $p = (0, 1, 0)$.
I can use the implicit function theorem with $g(x, y, z) = x^2 + y^2 +z^2$ to show that $M$ is a submanifold of $\mathbb{R}^3$ (it is an ellipsoid).
I have also found $dg = \begin{bmatrix} 2x & 2y & 2z\end{bmatrix}$, $dg_p = \begin{bmatrix}0 & 2 & 0 \end{bmatrix}$, $\ker(dg_p) = \{(x, 0, z) : x, z, \in \mathbb{R}\} = T_p\mathbb{R}^3$.
I want to use the invertible function theorem to show $f$ is a local diffeomorphism at $p$, but I'm confused about what exactly I need to show.
Obviously I need to show $df_p$ is bijective, and I found the matrix for $df$ as $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}$.
This is where I am stuck, how do I show this is bijective at $p$, and where does the submanifold come into it?
Let us use stereographic projection to show that $f$ is a local diffeomorphism.
Let $h_S : (x,y,z)\in M\setminus\{N\} \mapsto \left(\frac{x}{1-z},\frac{y}{1-z} \right)\in \mathbb{R}^2 $ be the stereographic projection with pole $S=(0,0,-1)$ the south pole. It is a diffeomorphism from $M \setminus \{N\}$ to $\mathbb{R}^2$, where $N =(0,0,1)$ is the north pole.
Let $g : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $g = f\circ {h_S}^{-1}$. If you can show that $g$ is a local diffeomorphism, the chain rule will show that $f$ is a local diffeomorphism at any point $p \in M \setminus \{N\}$. This is because $$ f = g\circ h_S $$ and thus, if $p=(x,y,z) \in M\setminus\{N\}$, one has $$ \mathrm{d}f_{p} = \mathrm{d}g_{h_S(p)}\circ \mathrm{d}{h_S}_{p} $$
The same study with $h_N$ the stereographic projection with pole $N$ will show that $f$ is a local diffeomorphism on $M\setminus \{S\}$ and the result will follow.