A variation on Ishaii's example is a closed embedding: its sheaf of relative differentials is 0, hence free of finite rank, even though it needn't be smooth.
However, k[e] / e^2 over k is not actually a counterexample (except in characteristic 2). The module of relative differentials of Spec k[e] / e^2 over Spec k is not free if the characteristic of k is not 2. Let A = k[e] and B = k[e] / e^2. Then
Omega(B) = Omega(A) (x) B / d(e^2) = k[e] / (e^2, 2e)
via the isomorphism Omega(A) --> A : dt --> 1. This is not isomorphic to B unless 2 = 0.
On the other hand, you can conclude that B is smooth if its cotangent complex is a vector bundle in degree 0. In the case of k[e] / e^2, the cotangent complex is
[ I(B/A) / I(B/A)^2 ---> Omega(A) (x) B ] = [ e^2 A / e^4 A ---> B de ]
in degrees [-1,0] and the differential is the universal derivation. (I write I(B/A) for the ideal of B in A.) Even in characteristic 2, the differential has a kernel, so the cotangent complex is not concentrated in degree 0.