Regular functions on a cusp with two points removed ($k[t^2,t^3{]}_{t^2(t-1)}=k[t,t^{-1},(t-1{)}^{-1}]$)

Parametrize the affine cuspidal cubic $x^2=y^3$ by $x=t^3,y=t^2$. Removing the zero set of $g=x-y$ removes exactly the parameter values zero and one. Its regular-function ring is the localization $k[t^2,t^3{]}_g$. To recover the normalization coordinate inside this ring, use $g^2=y(y^2-2x+y)$: inverting $g$ makes $y$ a unit, so $t=x/y$ lies in the localization. As $g=t^2(t-1)$ is a unit, so are $t$ and $t-1$, proving the displayed equality. This calculation is valid in every characteristic. At the smooth point $(1,1)$ of the original curve the local ring is instead $k[t{]}_{(t-1)}$: $y$ is a unit there, and every polynomial in $t$ nonzero at one is invertible. That point is absent from the displayed open subset, so its local ring is computed in the original curve.