Let $f$ be a linear functional where $f$ is not identically $0$. If $f(AB)=f(BA)$ for all $n \times n$ matrices $A$ and $B$, then $f(C)=0$ implies $C = AB-BA$ for some $A,B$.
I am not sure how to show that these are the only matrices in the kernel of $f$. Constructing $A$ and $B$ explicitly seems too difficult and trying to prove the contrapositive isn't getting me anywhere. Could I have a small hint?
The essential point to show is that the set $S$ of all commutators $AB-BA$ is a subspace of the space of square matrices, and that it has codimension$~1$; then any nonzero linear form that vanishes on all of $S$ necessarily vanishes only on$~S$. Since one can take for $f$ the trace function, the required property is certainly not vacuously true, and in can only hold if $S$ is equal to the set of traceless matrices (the kernel of the trace function). So to prove the statement is equivalent to proving that any traceless matrix is in$~S$.
This is indeed a fact, as was indicated in a link provided in a comment by totoro. It is not hard to see that commutators span the space of traceless matrices, but since there is no obvious way to write a linear combination of unrelated commutators as another commutator, there appears to be no shortcut to explicitly showing each traceless matrix to be a commutator; I can give no easy proof.