I'm having trouble calculating the image of a point $P=[0:0:1]$ under a polynomial map:$f:V \to \mathbb{P}^1 : [x:y:z] \mapsto [y:x]$ where $V = \left\{ [x:y:z] \in \mathbb{P}^2 ~\mid~ zy^2 = zx^2 + x^3 \right\}.$
What is $f([0:0:1])$?
Dead end
Notice that the given polynomials don't work at this point (but they do at every other point of $V$) since $[0:0]$ is not a valid expression for a point in $\mathbb{P}^1$. I found a different expression, $f([x:y:z]) = [zx+x^2:zy],$ but this expression is not defined at $[0:0:1]$, $[0:1:0]$, or $[-1:0:1]$, so it is an objectively worse expression. How do I find a third expression?
Similar problem where things work:
Let $\mathbb{P}^n$ be projective space over an algebraically closed field $k$, so that $\mathbb{P}^n$ consists of all $(n+1)$-tuples $[x_0 : x_1 : \cdots : x_n ]$ with $(x_0,x_1,\ldots,x_n) \neq (0,0,\ldots,0)$ with two tuples $[x_0 : x_1 : \cdots : x_n ] = [y_0 : y_1 : \cdots : y_n ]$ considered equivalent if there is some nonzero $\lambda$ in $k$ with $\lambda x_i = y_i$ for $i=0,1,\ldots,n$.
Let $V = \left\{ [x:y:z] \in \mathbb{P}^2 ~\mid~ x^2 + y^2 = z^2, (x,y,z) \neq(0,0,0) \right\}$ be the (projective) circle, and let $W = \left\{ [ x:y ] \in \mathbb{P}^1 ~\mid~ (x,y) \neq 0 \right\}$ be the projective line, and let $f : V \to W : [x:y:z] \mapsto \begin{cases} [ y-z : x ] & \text{ unless } [x:y:z] = [0:1:1] \\\ [ x : y+z ] & \text{ unless } [x:y:z] = [0:-1:1] \\ \end{cases}.$
Note that this rule for $f$ is globally valid and the function is well-defined. In other words, the two definitions agree on their overlap, and every point of $V$ is covered by one of the definitions.