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.
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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.