I am stuck on the following question. Sorry about the bad latex skills. Not sure what went wrong.
Is the map $f:S^n \rightarrow S^n$ orientation preserving? I constructed an atlas on $S^n$ consisting of two charts. They are $\{\phi_\pm, U_\pm, \mathbb{R}\}$ Where $U_\pm = S^n\backslash \{(0, ..., 0, \mp 1 )\}$ and $$\phi_\pm(x_1, ..., x_n, x_{n + 1}) = \frac{1}{1 \pm x^{n + 1}}(x_1, ..., x_n)$$. Then we have that the transition map $\phi_{-+}$ is given by $$\phi_{-+}(y_1, ..., y_n) = \frac{1}{y_1^2 + ... + y_n^2}(y_1, ..., y_n)$$.
I am not sure how to compute the Jacobian of this map to show whether its det is $\pm$.
You calculated the transition map $\phi_{-+}:\mathbb{R}^n\to\mathbb{R}^n$ correctly. The Differential $D_\phi$ is represented by the usual Jacobian consisting of the partial derivatives of the $i$-th component of $\phi_{-+}$ in direction of the $j$-th coordinate, i.e. $$D_\phi(x)=\left(\frac{\partial\phi_{-+}^i}{\partial y^j}\right)_{i,j=1,...,n}.$$
The sign of the Jacobian's determinant is in this case (since there is only one transition map) its orientation.