How to compute the differential of $f:S^2 \rightarrow S^2, z \mapsto 1/z$ at the north and south pole? 
Consider the identification $\mathbb{C} \cup \{ \infty \} \simeq S^2$. Show that $f:S^2 \rightarrow S^2, z \mapsto 1/z$ is a $\mathcal{C}^\infty$ function and compute its differentials in the north and south poles.

I understand (see for example this question) that we may view this map as a rotation of $S^2$ by 180 degrees about the diameter with endpoints $(-1/2,0,1/2)$ and $(1/2,0,1/2)$ and that it may be represented by the matrix
$$f(x,y,z) = \begin{pmatrix} x \\ -y \\-z \end{pmatrix} = \begin{pmatrix} 1 &0 & 0 \\ 0 &-1 & 0 \\ 0 &0 &-1 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}$$
and we know that linear maps are smooth.
Now to the differential. If I am not mistaken the differential can be computed via the Jacobi matrix.
However, then we would have
$$Df=\begin{pmatrix} 1 &0 & 0 \\ 0 &-1 & 0 \\ 0 &0 &-1 \end{pmatrix}$$
, but I do not think that this is correct. Could you please tell me what I am doing wrong?
 A: In the linked question the sphere $S^2$ has center $(0,0,1/2)$ and radius $1/2$. A more standard interpretation is to regard it as
$$S^2 = \{ p \in \mathbb R^3 \mid \lVert p \rVert = 1 \}$$
which is centered at the origin and radius $1$.
Stereographic projection from the north pole is given by the same formula as in Rotation around the diameter in Riemann sphere. Using the same approach as in its answer one can see that the function $1/ z : \mathbb C \to \mathbb C$ corresponds to a rotation of $S^2$ around the $x$-axis by an angle of $180^o$.
This rotation is a linear automorphism $R : \mathbb R^3 \to \mathbb R^3$ which is described by the matrix $$M_R =\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0& -1  
 \end{pmatrix} .$$
Its differential has indeed Jacobian matrix $JR\mid_p = M_R$ at all $p \in \mathbb R^3$.
$R$ restricts to a smooth map $r : S^2 \to S^2$ which rotates $S^2$ around the $x$-axis by $180^o$. Its differential at the north pole $n$ is $Dr \mid_n : T_nS^2 \to T_sS^2$ from the tangent space $T_nS^2$ at $n$ to the tangent space $T_sS^2$ at the south pole $s$ .
Both tangent spaces are $\bar{\mathbb R}^2 = \mathbb R^2 \times \{0\}$ = $x$-$y$-plane in $\mathbb R^3$. We have the following commutative diagram:
$\require{AMScd}$
\begin{CD}
\bar{\mathbb R}^2 = T_nS^2 @>{Dr \mid_n}>> T_sS^2 = \bar{\mathbb R}^2\\
@V{}VV @VV{}V \\
\mathbb R^3 = T_n\mathbb R^3 @>>{DR \mid_n}> T_s\mathbb R^3 = \mathbb R^3\end{CD}
Here the vertical arrows are inclusion maps.
This shows that
$$Dr \mid_n(x,y,0) = (x,-y,0) .$$
The same is true for $Dr \mid_s : T_sS^2 \to T_nS^2$.
Both maps are reflections in the $x$-axis of $\bar{\mathbb R}^2$.
