I have got stuck trying to solve the following problem. Let $X=-zx \frac{\partial}{\partial x} -zy \frac{\partial}{\partial y} + (1-z^2) \frac{\partial}{\partial z}$ be a vector field in $\mathbb{R}^3$. I am asked to determine the expression of the restriction of $X$ to the sphere by means of the following chart $\phi: (u,v) \to (\frac{x}{1-z}, \frac{y}{1-z})$.
What I tried was to express $X|_{S^2}$ as $A_u \frac{\partial}{\partial u} + A_v \frac{\partial}{\partial v }$ and then try to determine the functions $A_u$ and $A_v$ using that
$$ \frac{\partial}{\partial u} = \frac{1}{1-z} \frac{\partial}{\partial x} + \frac{x}{(1-z)^2} \frac{\partial}{\partial z}$$
$$ \frac{\partial}{\partial v} = \frac{1}{1-z} \frac{\partial}{\partial y} + \frac{y}{(1-z)^2} \frac{\partial}{\partial z}$$
But I don't get anything that makes sense. Where I've gone wrong? How can I solve this?
It's better to express $\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}$ in terms of $\frac{\partial }{\partial u}$, $\frac{\partial }{\partial v}$:
$$\frac{\partial }{\partial x}=\frac{1}{1-z}\frac{\partial }{\partial u}\\ \frac{\partial }{\partial y}=\frac{1}{1-z}\frac{\partial }{\partial v}\\ \frac{\partial }{\partial z}=\frac{x}{(1-z)^2}\frac{\partial }{\partial u}+\frac{y}{(1-z)^2}\frac{\partial }{\partial v}$$
Now you can plug this into the original expression and simplify.