Using the Riemann Hurwitz Formula I am working with the function $f(z)=\frac{z^3}{1-z^2}$ from the Riemann Sphere to itself. I'm trying to show that this satisfies the Riemann-Hurwitz formula given 
$$2g(X)-2=\deg(F)(2g(Y)-2)+\sum_{p}(\text{mult}_p(f)-1)$$
The genus of domain and range are the same and equal to $0$. The function $f$ is a degree three map because exactly three things map to $\infty$. Only $p=0$ has multiplicity $3$ and this is the only point with multiplicity greater than one. But then I get Riemann-Hurwitz as $$-2= 3(-2)+(3-1)$$
 which is basically $-2=-4$ which is utter rubbish. What am I doing wrong? :(
 A: What you're seeing is that there must be other ramification that you haven't noticed. In fact, from a representation theory of the symmetric group (don't worry if you don't know what that is), you can show that it isn't possible for there to be a map from $\mathbb{P}^1 \to \mathbb{P}^1$ that is ramified at only one point.
(The short description: ramification points correspond to permutations, and with regards to $\mathbb{P}^1$, we are talking about how many ways we can factor a given permutation (say, the one over 0) into cycle types given by the ramification. However, if there is only one permutation over 0, then we are trying to factor the permutation $(1\, 2\, 3)$ into... no permutations.)
Anyhow, you can see that there are other ramification by looking at the derivative. It will be zero exactly at any ramification point, and its order of vanishing at the point is the ramification index. In your case, this is given by
$$
f'(z) = z^2\frac{3-z^2}{(1-z^2)^2}
$$
and so you can see that, in addition to being ramified at 0, this is also simply ramified at $\pm \sqrt{3}$.
