Smoothness depends on parametrization

Question

Does the smoothness (meaning infinitely differentiable) of a function depend on its parametrization? Suppose we have the function $f(t) = [t^2,t^{\frac{1}{3}}]^T$ on [0,1]. Then $\nabla f(t) = [2t,\frac{1}{3}t^{-2/3}]^T$. But $t^{\frac{1}{3}}$ is not differentiable at t=0. If I just replaced $t$ with $x^3$, would this make the function smooth?