TITLE: $SL(2,\mathbb{C})$ action on $\mathbb{C} P^1$ induces action of Lie algebra by vector fields
QUESTION [5 upvotes]: I am in need of some assistance to solve problem 3.8 in Kirillov's An Introduction to Lie Groups and Lie Algebras. Copied below:

Let $SL(2, \mathbb{C})$ act on $\mathbb{C} P^1$ in the usual way:
  $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} [x:y] = [ax+by:cx+dy].$$ 
  This defines an action of $\mathfrak{g} = \mathfrak{sl}(2,\mathbb{C})$ 
  by vector fields on $\mathbb{C} P^1$. Write explicitly
  the vector fields corresponding to $h, e, f$ in terms of the
  coordinate $t = x/y$ on the open cell $\mathbb{C} \subset \mathbb{C} P^1$.

I am aware that the action of $SL(2,\mathbb{C})$ on $\mathbb{C}P^1$ is equivalent to a map $\rho: SL(2,\mathbb{C}) \to \mathbb{C}P^1$, and further the pushforward $\rho_*$  defines the action of the Lie algebra. In terms of one-parameter subgroups, recalling $$ h = \begin{pmatrix}1&0\\ 0&-1\end{pmatrix},$$ $\exp(sh)$ is a curve in the Lie group with tangent vector $h$ at $s=0$, and so the action of $h$ on $\mathbb{C}P^1$ is defined to be  $$ \frac{d}{ds}\exp(sh)([x:y])|_{s=0} = \frac{d}{ds} [e^s x: e^{-s} y]|_{s=0}$$ at which point I'm a little confused. 
Edit: I should also add that this is a homework assignment, so hints are strongly preferred over full solutions.

REPLY [2 votes]: You already have most of the solution. Now you just need to write the expression you got in means of the desired coordinate.
The complex coordinate on the set $U_y:=\{[x:y]|y\neq0\}$ is 
$$
\varphi:U_y\to\mathbb{C},\quad[x:y]\mapsto\frac{x}{y}.
$$
This means that the path whose velocity you are after is given by
$$
\gamma:(-\epsilon,\epsilon)\to\mathbb{C},\quad s\mapsto[e^sx:e^{-s}y]\mapsto e^{2s}\frac{x}{y}.
$$
Can you take it from here?