Here is my problem:
Let $G=\langle a,b|a^l=b^3=1,(ab)^3=(a^{-1}b)^3 \rangle$. Find the order of $\frac{G}{G'}$ and then verify that if $G$ is metabelian.
What I have done: I added the relation $[a,b]=1$ to $G$'s relations and found that:
$\frac{G}{G'}=\langle a,b|a=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3$
$\frac{G}{G'}=\langle a,b|a^2=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_6$
$\frac{G}{G'}=\langle a,b|a^3=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3\times\mathbb Z_3$
$\frac{G}{G'}=\langle a,b|a^6=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3\times\mathbb Z_6$
for the last part of the question, I have to probe if $G'$ is abelian or not. This is my idea:
For 1. ; Since $a=1$ is a relation of the quotient group so, $a\in G'$ and also $b^3\in G'$. This means that any conjugations of $a$ or $b^3$ would be in the derived subgroup as well. For example $b^{-1}ab$ or $a^{-1}b^3a$. Now I think to use such these conjugations and the relations of $G$ simultaneously, to find a presentation of $G'$ and see that if this subgroup is abelian.
This would be really a long harsh road :) solving the problem just in the case 1. Honestly, I don't know any way but Todd-Coxeter Algorithm for such these problems. May I ask to note me some useful hint? Thanks