Condition for ramification points of a projection $\pi : X \to \mathbb{P}^1 $ Let $X = \{F=0\} \subset \mathbb{P}^2$ be a projective plane curve and let $\pi : X \to \mathbb{P}^1$ be defined by $\pi [x:y:z] \to [x:y]$.
I'm trying to understand why the following is true: 

The holomorphic map $\pi$ is ramified at $p \in X$ Iff $\frac{\partial F}{\partial z} (p)=0.$

So far I understand the case where $X$ is an affine curve in $\mathbb{C}^2$ and the map $\pi$ projects to $\mathbb{C}$. Here the same condition condition applies as i've been able to prove. 
A projective plane curve $X$ is locally just an affine curve so the same condition must apply but the projection here is confusing since there are two coordinates. Could someone carefully outline the argument needed here to conclude the projective case from the affine case?
 A: You better assume  that the curve $X$ is smooth: else the morphism $\pi$ might not be defined everywhere and moreover ramification would not be a very clear concept.
Your morphism $\pi:X\to \mathbb P^1$ is best seen as the projection of the curve $X$ from the point $O=(0:0:1)$      to the line at infinity $z=0$.
Indeed that line at infinity  is a copy of $\mathbb P^1$ with coordinates $(x:y:0)$ .
Now, the morphism $\pi$ is ramified at $P=(a:b:c)\in X$ iff the line $\overline {OP}$ is tangent to $X$ at $P$.
But the tangent line  at $P$ to $X$ (remember that $X$ is assumed smooth) has equation $$\frac{\partial F}{  \partial x}(P)\cdot x+  \frac{\partial F}{  \partial y}(P)\cdot y+\frac{\partial F}{  \partial z}(P)  \cdot z=0      $$ 
That tangent line will pass through $O=(0:0:1)$ iff $\frac{\partial F}{  \partial x}(P)\cdot 0+  \frac{\partial F}{  \partial y}(P)\cdot 0+\frac{\partial F}{  \partial z}(P)  \cdot 1=0 $ i.e. iff  $$\frac{\partial F}{  \partial z}(P)=0$$ which is thus the required condition for $P\in X$ to be a ramification point for $\pi$.
[Notice the little trick of replacing the condition the line $\overline {OP}$ is tangent to $X$ by the condition the tangent line to $X$ at $P$ goes through $O$ ]
