I've been study the proof of the genus formula and I am a little confused about the ramification index of one of the maps involved. Let $C$ be a projective algebraic curve in $\mathbb{C}P^2$ such that $[0,1,0] \notin C$. Define $$\varphi: C \to \mathbb{C}P^1 $$ $$\varphi([x,y,z]) = [x,z] $$ Then $\varphi$ is a holomorphic map. I want to show that the ramification index $\varphi$ at $[a,b,c] \in C$ is the degree of the zero of the map $y \mapsto P(a,y,c)$ at $b$. The definition I know of ramification index is the smallest number $n$ such that the $n$-th derivative of the map (in a local chart) is non-zero at the point.
Take $p=[a,b,c]\in C$. Since $[0,1,0] \notin C$, we know that not both of $a,c$ are zero. Say $c\not = 0$. By Euler's relation, we know that at least one of $\partial P/\partial x,\partial P/\partial y$ is non-zero at $[a,b,c]$. The case where $\partial P/\partial y \not = 0$ is easy, so say $\partial P/\partial x \not = 0$. Then by the implicit function theorem we may write $y$ as a function of $x$. Since $c\not =0$, using the usual local coordinate on $\mathbb{C}P^1$ gives that our map $\varphi$ is given by $x \mapsto y(x)$ in local coordinates.
I'm not sure how to show that $y(x)$ has derivatives at $a$ as required. Elementary arguments would be greatly appreciated.