Rational locus of a function defined on $x^2+x^3=y^2$

Question

We have a curve $X$ on $\mathbb{A}^2$ given by $y^2=x^2+x^3$. Consider the rational function $f$ on $X$ which maps $(x,y)\in X$ to $\frac{y}{x}$. There is a nice geometric interpretation of $f$: if we draw a line $l=\{x=1\}$ then $(1,f(x,y))$ is the intersection of $l$ with the line that contains $0$ and $(x,y)$. However this 'geometric' map is well defined only when $(x,y)\ne 0$. I'd like to ask if it means that $f\in \mathbb{k}(X)$ also cannot be defined at $(0,0)$? So, is it true that $(0,0)$ is in rational locus of $f$?

Answer 1

To see that $f$ cannot be extended to the point $(0,0)$, one can argue as follows, at least if the ground field is the complex numbers.

For $(x,y) \neq (0,0)$, $f(x,y)$ is the slope of the chord joining $(0,0)$ to $(x,y)$. But at $(0,0)$ the curve has two branches with distinct tangent directions, so the value of $f(x,y)$ approaches two different limits as we approach $(0,0)$ in these different directions.