First, this is the definition of regular surface I'm using.
A subset $S\subset\mathbb{R}^3$ is a regular surface if, for each $p\in S$, there exist a neighborhood $V\subset\mathbb{R}^3$ and a map $x:U\to V\cap S$ where $U$ is an open subset of $\mathbb{R}^2$ such that $x$ is differentiable ($C^\infty$), an homeomorphism and its differential, $dx_q:\mathbb{R}^2\to\mathbb{R}^3$ injective.
Then, we define differentiability of a continuous function between two surfaces as follows:
$\varphi: V_1\subset S_1\to S_2$ where $V_1$ is an open subset of the regular surface $S_1$ to the regular surface $S_2$, $\varphi$ is differentiable at $p\in V_1$ if given two local maps $x_1:U_1\to S_1$ and $x_2:U_2\to S_2$ with $p\in x_1(U_1)\subset x_2(U_2)$, the map $x_2^{-1}\circ\varphi\circ x_1:U_1\to U_2$ is differentiable at $q=x^{-1}(p)$.
After these definitions, I have the following example which I don't fully understand and seems pretty obvious:
Let $x:U\subset\mathbb{R}^2\to S$ a local map ($S$ is a regular surface). For each $p\in x(U)$ and every local map $y:V\subset\mathbb{R}^2\to S$ at $p$, we have $$x^{-1}\circ y:y^{-1}(W)\to x^{-1}(W),\text{ where }W=x(U)\cap x(V)$$ is differentiable, and then, $U$ and $x(U)$ are diffeomorphic.
Why is $x^{-1}\circ y$ differentiable? I know $x$ and $y$ are differentiable, bijective, but in order to know $x^{-1}\circ y$ is differentiable wouldn't we need $x^{-1}$ also differentiable? I feel there is something I'm missing.
Edit 1: I accidently skipped some notes about change of parameters, that might be the answer.
Edit 2: Indeed, change of parameters has non-vanishing jacobian and hence it's an isomorphism, which implies by Inverse Function Theorem that it is a local diffeomorphism.