Determining a derivation on the unit sphere of the $\mathbb{R}^3$ Let $S^2 \subseteq \mathbb{R}^3$ be the unit sphere in $\mathbb{R}^3$ (which is a smooth manifold of dimension $2$). Let $\phi = (x_1, x_2): S^2 \backslash \{N\} \to \mathbb{R}^2 $ be the stereographic projection of the $S^n$, where $N = e_3 = \pmatrix{0 \\ 0 \\1}$ is the north pole, i.e. $\phi$ is given by 
$$\phi(u) = \phi(u_1, u_2, u_3) = \left( \frac{u_1}{1 - u_3}, \frac{u_2}{1 - u_3} \right) $$
Now let $f: \mathbb{R}^3 \to \mathbb{R}$ be given by $f(u) = u_3$, i.e. $f$ is the third coordinate projection.
For any $p \in S^2$, I now want to determine the derivation $\frac{\partial}{\partial x_i} \mid_p (\overline{f})$ (where $\overline{f}$ denotes the function germ of $f$).
I must admit that I haven't yet fully understood how to determine a derivation like this. So do I simply need to determine the partial derivatives of $f$, and then determine $\phi(p)$ and sum up the partial derivatives of $f$ correctly to get the derivative of $f$ in the "direction" $\phi(p)$? I'm not really sure if that's the right approach, though.
 A: Here is how you determine
$$
\frac{\partial}{\partial x_i}|_p f
$$
in general for a smooth function $f$ defined on some neighborhood of $p$ in $S^2$. (The vector fields ${\partial}/{\partial x_i}$ are only defined on $S^2\setminus\{N\}$, so we have to assume $p\neq N$ for all this to make sense.) We may assume that $f=f(x,y,z)$ is a smooth function defined on a neighborhood of $p$ in $\mathbf R^3$. Note that the inverse $\phi^{-1}$ of the stereographic projection $\phi$ is defined by
$$
\phi^{-1}(x_1,x_2)=\left(\frac{2x_1}{||x||^2+1},\frac{2x_2}{||x||^2+1},\frac{||x||^2-1}{||x||^2+1}\right),
$$
where $||x||^2=x_1^2+x_2^2$. Then
$$
\frac{\partial}{\partial x_i}|_p f=\frac{\partial (f\circ\phi^{-1})}{\partial x_i}(\phi(p)).
$$
We compute the partial derivative using the chain rule. For example for $i=1$ this gives:
$$
\begin{multline*}
\frac{\partial (f\circ\phi^{-1})}{\partial x_1}=
\frac{\partial f}{\partial x}\cdot \frac{\partial \phi_1^{-1}}{\partial x_1}+
\frac{\partial f}{\partial y}\cdot \frac{\partial \phi_2^{-1}}{\partial x_1}+
\frac{\partial f}{\partial z}\cdot \frac{\partial \phi_3^{-1}}{\partial x_1}=\\
\frac{\partial f}{\partial x}\cdot \frac{2(||x||^2+1)-4x_1^2}{(||x||^2+1)^2}+
\frac{\partial f}{\partial y}\cdot \frac{-4x_1x_2}{(||x||^2+1)^2}+
\frac{\partial f}{\partial z}\cdot \frac{2x_1(||x||^2+1)-(||x||^2-1)2x_1}{(||x||^2+1)^2}=\\
\frac{\partial f}{\partial x}\cdot \frac{2-2x_1^2+2x_2^2}{(||x||^2+1)^2}+
\frac{\partial f}{\partial y}\cdot \frac{-4x_1x_2}{(||x||^2+1)^2}+
\frac{\partial f}{\partial z}\cdot \frac{4x_1}{(||x||^2+1)^2},
\end{multline*}$$
the first factor of each term being evaluated at $p$, the second at $\phi(p)$.
Doing the latter evaluation gives an expression in terms 
of the coordinate functions $x,y,z$ on $S^2$ only:
$$
\begin{multline*}
\frac{\partial}{\partial x_1}|_p f=
\frac{\partial f}{\partial x}\cdot \frac{2-2x^2/(1-z)^2+2y^2/(1-z)^2}{(x^2/(1-z)^2+y^2/(1-z)^2 +1)^2}+
\frac{\partial f}{\partial y}\cdot \frac{-4xy/(1-z)^2}{(x^2/(1-z)^2+y^2/(1-z)^2 +1)^2}+
\frac{\partial f}{\partial z}\cdot \frac{4x/(1-z)}{(x^2/(1-z)^2+y^2/(1-z)^2 +1)^2}=\\
\frac{\partial f}{\partial x}\cdot \frac{(1-z)^2(2(1-z)^2-2x^2+2y^2)}{(x^2+y^2+(1-z)^2)^2}+
\frac{\partial f}{\partial y}\cdot \frac{-4xy(1-z)^2}{(x^2+y^2+(1-z)^2)^2}+
\frac{\partial f}{\partial z}\cdot \frac{4x(1-z)^3}{(x^2+y^2+(1-z)^2)^2}=\\
\frac{\partial f}{\partial x}\cdot \frac{4(1-z)^2(1-z-x^2)}{4(1-z)^2}+
\frac{\partial f}{\partial y}\cdot \frac{-4xy(1-z)^2}{4(1-z)^2}+
\frac{\partial f}{\partial z}\cdot \frac{4x(1-z)^3}{4(1-z)^2}=\\
(1-z-x^2)\cdot\frac{\partial f}{\partial x}-
xy\cdot\frac{\partial f}{\partial y}+
x(1-z)\cdot\frac{\partial f}{\partial z},
\end{multline*}$$
which is the formula we where after.
It is interesting to note that it follows that the vector field $\partial/\partial x_1$ has smooth continuation over the whole sphere $S^2$, and that it vanishes at $N$.
Anyway, for your function $f(x,y,x)=z$, one now obtains
$$
\frac{\partial}{\partial x_1}|_p f=x(1-z).
$$
By symmetry, one has
$$
\frac{\partial}{\partial x_2}|_p f=y(1-z).
$$
