Is the cuspidal cubic $\{y^2 = x^3\} \subset \Bbb R^2$ not smooth?

The cuspidal cubic $C := \{(x, y) : y^2 = x^3\} \subset \Bbb R^2$ certainly admits a smooth structure: The projection $C \to \Bbb R$ onto the $y$-axis is continuous and has continuous inverse $y \mapsto (y^{2 / 3}, y)$, so it is a homeomorphism, and hence using it to pull back the (usual) smooth structure on $\Bbb R$ defines a smooth structure on $C$.

Like you say, however, $C$ is not an embedded (regular) submanifold of $\Bbb R^2$: the pullback by inclusion $C \hookrightarrow \Bbb R^2$ does not define a smooth structure on $C$: If we regard tangent vectors as equivalence classes of differentiable curves into $C$, we can observe that any differentiable curve $\gamma: (-\epsilon, \epsilon) \to C$ such that $\gamma(0) = (0, 0)$ must satisfy $\gamma'(0) = (0, 0)$.