 Solution

Work over the usual algebraically closed ground field of characteristic zero. Away from $x=0$, a line of slope $t=y/x$ through the singular point meets the curve with $t^2=x-1$. ThusSubstitution verifies the defining equation. On $x\ne 0$ and $t^2+1\ne 0$, the two maps are inverse morphisms, so this is a birational map to ${\mathbb{A}}^1$. The two parameters $t=\pm i$ both map to the node $(0,0)$, explaining why the globally defined parametrization is not a global isomorphism.