Suppose that for a submanifold $H$ of $\mathbb{R}^3$ we have two charts
$$\varphi_1:U=\{(x,y,z)\in\mathbb{R}^3:x\neq0\}\rightarrow\varphi_1(U)$$
$$\varphi_2:U=\{(x,y,z)\in\mathbb{R}^3:y\neq0\}\rightarrow\varphi_2(U).$$
s.t. $\varphi_1(x,y,z)=(y,z)$ and $\varphi_2(x,y,z)=(x,z)$.
Let $f:H\rightarrow\mathbb{R}^2$ be the morphism $f(x,y,z)=(x,z)$.
Compute $(df)_a$ in those charts.
My attempt (for $\varphi_1$):
I know that $$(df)_a=\sum_i \frac{\partial \phi}{\partial x_i}(\varphi(a))(dx_i)_a$$
where $f \circ \varphi^{-1} = \phi$.
My question is: what is the expression of $\varphi_1^{-1}$, in order to get $\phi$? Maybe it is a silly question but I'm not used to these problems.
Thanks in advance.
You can cover the one-sheeted hyperboloid $$ H = \left\{ (x,y,z) \in \Bbb{R}^3 : x^2 + y^2 = 1 + z^2 \right\} $$ with the four charts $(U_x^{\pm},\varphi_x^{\pm}),(U_y^{\pm},\varphi_y^{\pm})$ where $$ \begin{align} U_x^{\pm} &:= \left\{ (x,y,z) \in H : x \gtrless 0 \right\} & \varphi_x^{\pm}(x,y,z) &= (y,z) \\ U_y^{\pm} &:= \left\{ (x,y,z) \in H : y \gtrless 0 \right\} & \varphi_y^{\pm}(x,y,z) &= (x,z) \end{align} $$
Now, observe that your map $f(x,y,z) = (x,z)$ coincides with $\varphi_y^{\pm}$ on $U_y^{\pm}$, so it is clearly differentiable on $U_y^{\pm}$. Furthermore, we have $$ (f \circ (\varphi_y^{\pm})^{-1}) (x,z) = (x,z) $$ so $J_a(f) = \bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$ for every $a \in U_y^{\pm}$.
The remaining points to check lie on a hyperbola $$ H_0 := H \setminus (U_y^+ \cup U_y^-) = \left\{ (x,y,z) \in H : y = 0 \right\} $$ Since $(\varphi_x^{\pm})^{-1} (y,z) = \left(\pm \sqrt{1 + z^2 - y^2},y,z\right)$ we have $$ (f \circ (\varphi_x^{\pm})^{-1})(y,z) = \left(\pm \sqrt{1 + z^2 - y^2}, z\right) $$ which is indeed differentiable in a small enough neighbourhood of $H_0$.
Note: Since this seems to be homework I'll leave the computation of the Jacobian (or equivalently, of the differential) to you. Drop a comment below if you need help with it.