Show that there is a smooth vector field on $\Bbb S^2$ that vanishes at exactly one point. [Hint: try using stereographic projection].
The idea is clear. The stereographic projection $\sigma:\Bbb S^2\setminus\{N\}\to\Bbb R^2$ by $(x,y,z)\mapsto \left({x\over 1-z},{y\over 1-z}\right)$ is a diffeomorphism and if we write $(u^1,u^2)$ as a coordinate of $\Bbb R^2$ then $\partial/\partial u^1$ is a nowhere vanishing vector field on $\Bbb R^2$. Hence, we pushforward $\partial/\partial u^1$ via $\sigma^{-1}$ which is $\sigma^{-1}_*(\partial/\partial u^1)$.
Then it's a nowhere vanishing vector field on $\Bbb S^2\setminus N$. Define the (rough) vector field $\tilde{X}$ on $\Bbb S^2$ by
$$\tilde{X}(p)=\begin{cases} ((\sigma^{-1})_*(\partial/\partial u^1))_p & p\neq N \\ 0 & p = N \end{cases}$$
I'm hoping this is a smooth vector field. To show this, I'm planning to compute the coordinate representation of $\tilde{X}$ near $N$. So consider another stereographic projection $\tilde{\sigma}:\Bbb S^2\setminus S\to \Bbb R^2$ by $(x,y,z)\mapsto \left({x\over 1+z},{y\over 1+z}\right)$. I'm trying to compute the local presentation of $\tilde{X}$ w.r.t. this coordinate $\tilde{\sigma}$ but don't know what to do next. Please help.