Show that the the group with presentation $$\langle x, y\ \mid\ x^2=y^2x^2y,\ (xy^2)^2=yx^2, \ yx^{-1}y^2=x^n\rangle $$ is cyclic of order $3(n+1)$, for $n=0 \mod 3$ or $n= 1 \mod 3$, $n\ge 0$.
This presentation is related to the group presented in problem 476854 and is also related to problem 876731.
By the same argument as for Group elements $x$ and $y$ satisfying $x^2 = y^2x^2y$ and $yx^{-1}y^2 = x^7$ commute. the first and third relations imply $yxy^{-1}=x^{n+2}$ and $y^3 = x^{2n+2}$. So $\langle y^3 \rangle \le Z(G)$.
As before, let $H= \langle y^3 \rangle$. If $n$ is odd, then the same argument as there gives $y^2xy^{-2}=x$ so $H$ is abelian, and $x^{n+1}=1$ in $H$. If $n$ is even then, since $y$ conjugates $x$ to an even power of itself, the order of $x$ in $H$ must be odd, so $x^{n+1}=1$ in $H$ and again $H$ is abelian. So in any case $H = \langle x \rangle \times \langle y \rangle = C_{n+1} \times C_3$. This is also true in $G/G'$, so $\langle y \rangle = G' \le Z(G)$. Note also that in the group $H$, the second relation is redundant.
Now, if $n \not\equiv 2 \mod 3$, then $H$ is cyclic, so $G/Z(G)$ is cyclic and hence $G$ is abelian, so $G'=1$ and $G \cong H$, which answers your question
What happens when $n \equiv 2 \mod 3$? Note that, since $y^3 \in Z(G)$, $[x,y]^3=[x,y^3]=1$, so $|G'| \le 3$. I found by computer calculations that, in the group defined by just the first and third relations, we have $|G'|=3$. But the second relation gives $xy^2xy^2=x^{2n+4}y^4 = yx^2=yx^{2n+4}$, so $y^3=1$, and hence $x^{2n+2} = x^{-(n+1)}=1$, so $x^{n+1}=1$, so the 3-relation group is also abelian in this case.