Hurwitz formula Well, I'm in distress over this problem for a week at least. I don't know if the poles and zeros are computed in the usual way, and no idea for the branch points apart from taking derivatives. Moreover the degree of the function seems to be one (because apparently the is only a simple pole and a simple zero) but I'm not sure. And I really don't how the curve may influence the function.
So, this is how the problem states:
Consider the curve $$C = \{[X,Y,Z] \in \mathbb{P^2C} | X^4 + Y^3Z - Z^3Y = 0\}$$
Consider over $C$ the meromorphic function $$g = \dfrac {Y}{Z}$$ compute poles and zeros of $g$. Find the branch points and verify Hurwitz formula.
Thank you for your help !
 A: Long sketchy hint:  In order to find the zeros of $g$ divide through the equation of the curve by $Z^4$ to get the affine equation
$$
\begin{align}
x^4+y^3 - y & =0 \\
x^4 = y-y^3 &= y\,(1-y^2)
\end{align}
$$
We see this has a zero at $(0,0)$ that corresponds to the point $(0:0:1)$ which we can take as a zero of $g$.  To find the poles of $g$ divide through the equation of the curve by $Y^4$ to get the affine equation
\begin{align}
x^4+z - z^3 & =0 \\
x^4 = z^3-z &= z\,(z^2-1)
\end{align}
And we see that there is a zero at $z=0$ that corresponds to the point $(0:1:0)$ which we take as the pole of $g$.
The ramification points occur where $x=0$ in either of the above equations, so we get the points each with ramification index $4$
\begin{align}
&(0:0:1) \\
&(0:1:1) \\
&(0:1:-1) \\
&(0:1:0) \\
\end{align}
giving $\{0,1,-1,\infty\}$ for the branch points.
By the degree genus formula $C$ has genus 3, so applying Riemann-Hurwitz viewing $g$ as a mapping from $C$ to the Riemann Sphere gives
$$ \begin{align}
2\cdot 3 -2 &=  4(2\cdot0 - 2) + \sum_{P} (e_P-1) \\
4 &= -8 + 4\cdot 3
\end{align}$$
