The problem: Let $V = V(y^2-x^2(x+1))$, and let $\overline{x}, \overline{y}$ denote the $I(V)$-residues of $x$ and $y$ in the coordinate ring $\Gamma(V)$. Set $z=\overline{y}/\overline{x}$. Find the pole sets for $z$ and $z^2$.
My progress: Since $I(V) = (y^2-x^2(x+1))$ (the polynomial $y^2-x^2(x+1)$ is irreducible), we have $\overline{y}^2 = \overline{x}^2(\overline{x} + \overline{1})$ in $\Gamma(V)$, so $z=\overline{y}/\overline{x} = (\overline{x}/\overline{y})(\overline{x}+\overline{1})$ and $z^2 = \overline{y}^2/\overline{x}^2 = \overline{x} + \overline{1}.$ Hence $z^2$ has no poles since one representation is a polynomial, and $z$ has no poles at $(x,y)$ if either $x\neq 0$ or $y\neq 0$. So I have concluded that the only possible pole for $z$ is $(0,0)$. However, I am unsure of how to check this point.
I would appreciate any hints.