 Solution

The parametrization $t\mapsto (t^3,t^2)$ maps ${\mathbb{A}}^1$ to the cusp. Off the origin its inverse is $t=x/y$, since $(x/y{)}^2=y$ and $(x/y{)}^3=x$ on $x^2=y^3$ with $y\ne 0$. This is an isomorphism between dense open subsets, so the cusp is birational to ${\mathbb{A}}^1$.

At the origin the differential of the defining polynomial is zero, giving a two-dimensional Zariski tangent space. The tangent space of ${\mathbb{A}}^1$ at every point is one-dimensional. An isomorphism induces tangent-space isomorphisms, so the cusp is not isomorphic to ${\mathbb{A}}^1$. The rational inverse cannot extend regularly through the singular point.