There is a claim saying that if both $G'/G''$ and $G''$ are cyclic groups, then $G''=1$, where $G'$ is the derived subgroup of the group $G$. I have been thinking of this by focusing the N/C Lemma to clear the problem for myself. I need a useful igniting hint(s). Furthermore, may I ask: are these kinds of groups well known? Of course, any group satisfying the above conditions will be metabelian and obviously is soluble.
This is theorem 9.4.2, page 146, in M. Hall's textbook on the Theory of Groups.
You are going in the right direction. It uses the N/C theorem, as in, the normalizer modulo the centralizer is a subgroup of the automorphism group.
Another hint: It is very similar to showing "If G/Z(G) is cyclic, then G is abelian.".
These groups were known as "metacyclic groups" for a few decades, though the name is now used slightly differently.
A special case where G′ and G/G′ have coprime order is very special: these are exactly the groups in which all Sylows are cyclic. They are also known as "Z-groups", though again the name means different things to different people.