Ramification points of projection to y-line Rick Miranda's "Algebraic Curves and Riemann Surfaces" chapter 2, lemma 4.6 says that:

Let $X$ be a smooth projective plane curve defined by a homogeneous polynomial $F(x,y,z) = 0$; consider the map $G: X \rightarrow \mathbb{P}^1$ defined by $G[x:y:z] = [x:z]$. Then $G$ is ramified at $p \in X$ if and only if $(\partial F/\partial y)(p) = 0$.

I understand that for point where either $x$ or $z$ is nonzero this theorem is true by looking at the affine parts and then considering local charts.  This only leaves the point [0:1:0] which might very well be a point in $X$.  But the map $G$ is not even defined there. Is there something missing here?
More generally, we know that for a curve $X$, ratio of two homogeneous polynomials $P,Q$ such that the denominator does not identically vanish on $X$ gives a meromorphic function, and therefore gives a holomorphic map to Riemann sphere $\mathbb{P}^1$ as $[P : Q]$. But for the points where both $P$ and $Q$ vanish, this map is not even defined?  What is going on?  Can I not write the  meromorphic function defined by $P/Q$ as $[P:Q]$ globally?  It seems to me that at points where both $P$ and $Q$ vanish I can simplify the function locally so that it either has a pole or a removable singularity. For a trivial example consider $P = Q = x$ and then $P/Q$  is just the constant function 1, but it still doesn't make sense to write $[x:x]$ because it won't be defined at points $x = 0$, assuming there is such a point on $X$.
 A: The result you're looking for here is that any rational map from a smooth curve $C$ to a projective variety $P$ extends uniquely to an honest morphism $C\to P$. This is known (in some places, at least) as the curve-to-projective extension theorem, and here's an outline of the proof: it suffices to prove the fact in the case when $C$ is affine and our map is defined on $C\setminus \{x\}$. Then our rational map $C\to P$ is specified by $c\mapsto [f_0(c):f_1(c):\cdots:f_n(c)]$ where the $f_i$ are rational functions on $C$. Since $\mathcal{O}_{C,x}$ is a DVR, we can divide through by an appropriate power of the uniformizer to arrange that none of the $f_i$ have a pole at $c$ and not all of the $f_i$ vanish at $c$, giving an honest map $C\to P$ extending our rational morphism. For a full accounting, see section 16.5 in Vakil's Foundations of Algebraic Geometry.

In the comments, some clarification regarding what happens in the case that $[0:1:0]$ is on our curve was requested. First, if $[0:1:0]$ is on $V(F)$, then $\frac{\partial F}{\partial y}([0:1:0]) =0$: if $\deg F = 1$, then $F$ is the linear equation $ax+bz$ for some $a,b\in k$; while if $\deg F \geq 2$ then every term in $F$ is divisible by either $x$ or $z$ and the same is true for $\frac{\partial F}{\partial y}$, so it must evaluate to zero.
Now I claim that the image of $[0:1:0]$ under the projection is $\left[\frac{\partial F}{\partial x}([0:1:0]):\frac{\partial F}{\partial z}([0:1:0])\right]$, where it is clear that the map on tangent spaces is zero, so the projection is ramified at $[0:1:0]$. Up to a linear change of coordinates preserving $[0:1:0]$, we may assume the tangent line to $C$ at $[0:1:0]$ is just $V(x)$. Dehomogenize by setting $y=1$: up to scaling, we can write $F=x(1+p(x))-zq(x,z)$ where $q$ has vanishing constant term. Then $$[x:z]=[x(1+p(x)):z(1+p(x))]= [zq(x,z):z(1+p(x))] = [q(x,z):1+p(x)]$$ which evaluates to $[0:1]$ when we set $x=z=0$. So the projection map is ramified at $[0:1:0]$ when that point is on our curve.
