Suppose $n \ge 3$ and $g,h \in S_{n}$ are such that the only non-fixed point of both $g$ and $h$ is $k$. I am trying to prove that the commutator of $g$ and $h$ is a 3-cycle with the only non-fixed points are $k, g(k)$, and $h(k)$. I think you would only need to look at four cases, computing each of these points under the commutator and one other point that is not any of these. So first, I used the convention that for any $\sigma, \tau \in S_{n}$, $\sigma\tau$ is the composition $\sigma \circ \tau$.
Now, assume $g(k) = i$ for some $i \neq k$ and $h(k) = j$ for some $j \neq k$. The first intuition is to apply the commutator to $k$: $ghg^{-1}h^{-1}(k)$.
I am not sure on how to proceed and deal with the inverses, but I think I am on the right track.