Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\dim X-\dim Y$?

**Edit:** As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{char}\,k=2$.