Difficulty calculating with rational functions 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.
 A: You can stop looking for a third expression for $f$: that rational function  is not defined at  $P$.
Note carefully that I'm not saying that $f$ has  a pole at $P$: I am saying that  it has a point of indeterminacy at $P$ and that $f$ can't be extended to a morphism $V\to \mathbb P^1$.
In contrast, a rational map from a smooth projective curve to a projective variety is actually a morphism i.e. it has no such indeterminacy: this explains your success with the circle .
However the point $P$ on $V$ is  singular : this explains your failure [or rather the failure of $f$ :-)] with the strophoid $V$ .  
And actually all this is visible with the naked eye in your example: in the affine part $z=1$ of the curve, the rational map $f$ sends the point $Q=(x,y)=[x:y:1]$ to the slope $f(Q)=y/x$ of the chord $\overline{PQ}$.
In the limit when $Q$ tends to $P$ that slope tends to $+1$ or $-1$ according to the branch of the strophoid on which $Q$ lies.
However the Zariski topology is too coarse to distinguish between these two branches and that is the geometric reason why $f$ is not defined at $P$.    
Edit
The last section above shows us the way to correct this failure of $f$.
We normalize the curve $V$ so that $P$ is replaced by two points $P_1, P_{-1}$ and $f$ will send these points respectively to $1$ and $-1$. 
Technically the normalization is  $$n:\mathbb P^1\to V:[u:v]\mapsto   [u^2v-v^3:u^3-uv^2:v^3]                    $$ and the composition of $n$ and $f$  is the everywhere defined morphism $$\tilde f=f\circ n:\mathbb P^1\to \mathbb P^1:[u:v]\mapsto [ u^3-uv^2:u^2v-v^3]=[u:v]$$
In other words $\tilde f$ is the identity!
Check that the points above $P=[0:0:1]$ are $P_1=[1:1], P_{-1}=[-1:1]$ and that $\tilde f(P_1)=[1:1]=\text {slope }(1/1), \;\tilde f(P_{-1})=[-1:1]=\text {slope } (-1/1)$
A: In order for $f=(f_0:\cdots:f_n)$ to define a rational map from some $X$ to $\mathbb P^n,$ we must have not all $f_i=0.$ This is part of the definition of a rational map, since otherwise we encounter the issue of mapping to $(0:\cdots:0),$ which makes no sense. So in your second example, you are exactly modifying the expression in order to define a rational map at every point of $V.$ In the first, you should say $f:[x:y:z]\to[y:x]$ unless if $x=y=0.$ As it stands, we have not assigned a value to $f((0:0:1)),$ and in order to do so, we must find an alternative expression for $f$ in a different chart (coinciding with the first expression on an overlap), as you did in the second example.
Variation on a theme:
Over the chart $z=1$ that we've been using, the blowup $\tilde V$ of $V$ can be computed as the strict transform of $V$ under $\mathrm{Bl}_{(0,0)}\mathbb A^2\subseteq\mathbb A^2\times\mathbb P^1\to\mathbb A^2.$ Away from $(0,0),$ the point $(x,y)$ corresponds to $(x,y;x:y),$ so if e.g. $x\neq 0$ we get $(x,y;1:y/x)$ as the unique point in the fibre above $(x,y).$ (As you noticed, this is the graph of your function!) But above the origin are two points, $(0,0;1:1)$ and $(0,0;1:-1)$ corresponding to the branches of your parametrization at $(0,0).$ There is a natural map $\tilde f:\tilde V\to\mathbb P^1,$ the restriction of the projection $\mathbb A^2\times\mathbb P^1\to\mathbb P^1,$ which lifts the original $f$ (and $\tilde f:\tilde V\to\mathbb P^1$ is essentially the identity, as with the normalization).
