This is my self-study exercise:
Let $G=\langle a,b\mid aba=b^2,bab=a^2\rangle$. Show that $G$ is not metabelian.
I know; I have to show that $G'$ is not an abelian subgroup. The index of $G'$ in $G$ is 3 and doing Todd-Coxeter Algorithm for finding any presentation of $G'$ is a long and tedious technique (honestly, I did it but not to end). Moreover GAP tells me that $|G|=24$. May I ask you if there is an emergency exit for this problem. Thanks for any hint. :)