Find the poles of a function over the rational function field. Let $V = V(y^{2} - x^2(x+1)) \subseteq \mathbb{A}^2$ and $\overline{x}, \overline{y} \in K(V)$. Let  $z = \dfrac{\overline{y}}{\overline{x}} \in K(V)$. Find the poles of $z$ and $z^2$.
I don't know even how to start this question. Can anyone help me?
This exercise is the 2.17 in fulton.
 A: It is essential for you to have the picture of the curve $y=\pm x\sqrt{1+x}\,$ in your mind or physically sitting before you.
Let me answer this in the language and from the viewpoint that I know. There are two points at the origin, $P^+$ and $P^-$, which you can see if you imagine lifting one branch of the curve up above the other. The point $P^+$ has the local uniformizing parameter $z-1$, and $P^-$ has the parameter $z+1$. The zeros of $y$ are the point $(-1,0)$ and the two points at the origin, while the only zeros of $x$ are the origin points. The poles are up at the single point “at infinity”, which has the local uniformizer $y/x=1/z$.
When I was teaching Calculus, I always gave this curve as an example of a parametrization,
\begin{align}
x&=t^2-1\\
y&=t^3-t\,,
\end{align}
and of course the parameter $t$ is your $z$. This parametrization shows that $K(V)$, defined to be the fraction field of $K[X,Y]\big/(Y^2-X^3-X^2)$, is equal to $K(z)$, and you see, looking at what $t=z$ does for you, where the (single) pole of $z$ is.
