Explicit computation for an elliptic curve. Let $C\subset \mathbb P^2$ be the smooth curve $Y^2Z=X^3+Z^3$ and let $p:C_0\to \mathbb A^1$ be the projection $(x,y)\mapsto x$ from the affine part $C_0$ of $C$ (described by $y^2=x^3+1$) onto the $x$ axis.
What is an explicit formula for the extension $P:C\to \mathbb P^1$ of $p$ to $C$, i.e. how do you write down $P$ explicitly in the neighborhood of the point at infinity $\infty _C=(0:1:0)$ of $C$ ?
Although I know quite well in theory how to deal with change of charts in projective space, I somehow seem  to make a mess of my computations with the coordinates $u=\frac XY, v=\frac ZY$ near $\infty_C$, in which $C$ has equation $v=u^3+v^3$.
Question:  Is it true that locally at $\infty _C$ we can write $P(u,v)=\frac{u^2}{1-v^2}$ ?
 A: Let me first say where the difficulty lies.  Although the map $(x, y) \mapsto x$ is defined on all of $\mathbb A^2$ (not just on $C_0$), its projective version $\pi:  [x : y : z] \mapsto [x : z]$ is undefined precisely at the point $\infty_C$ in question.  Although we know $\pi|_{C_0}$ can be extended uniquely to a map $\widetilde{\pi}:  C \to \mathbb P^1$ (like any map from a curve to a projective variety), $\pi$ itself cannot be extended across $\infty_C$.  So we do need to use the equation for the curve somehow.
To find the extension, first notice that the equation $v = u^3 + v^3$ on your second affine chart implies that $v$ vanishes to order $3$ at $\infty_C$, while $u$ vanishes only to order 1.  In particular, in order to make sense of the map $(u, v) = [x : 1 : z] \mapsto [x : z]$ at $x = z = 0$, we must divide the target coordinates by $x$.  Let's try to manipulate the coordinates in a way that allows us to do so:
\begin{align*}
[x : 1 : z] & \mapsto [x : z] & \text{when } (x, z) \neq (0, 0) \\
& = [x(1-z^2) : z(1-z^2)] & \text{when additionally } z \neq \pm 1 \\
& = [x(1-z^2) : x^3] \\
& = [1-z^2 : x^2],
\end{align*}
which doesn't vanish at $\infty_C$.  (In fact it's well-defined on the affine chart with $y = 1$, except at $[0 : 1 : \pm 1]$.)  So $\widetilde{\pi}$ must send $(u, v) = [x : 1 : z]$ to $[1-z^2 : x^2]$ on this locus.  And yes, the image point can be interpreted as $\frac{u^2}{1-v^2}$ in the affine patch of $\mathbb P^1$ "at infinity".
