= Solution
The <cyclic subgroup> $\langle x^3\rangle$ is normal: <conjugation> by $y$ sends $x^3$ to $x^{12}=x^3$, and <conjugation> by $x$ fixes it. Modulo this <subgroup>, the relation $yxy^{-1}=x^4$ becomes $yxy^{-1}=x$, so the <quotient group> is <Abelian>. Thus the <commutator subgroup> $G'$ is contained in $\langle x^3\rangle$.
Conversely the displayed relation gives the <commutator> $yxy^{-1}x^{-1}=x^3$, so $\langle x^3\rangle\subseteq G'$. Hence
$$
G'=\langle x^3\rangle\cong C_3,\qquad G''=\{e\}.
$$
Since $x$ has <element order> $9$, $x^3\ne e$, so the first <commutator subgroup> is nontrivial. Therefore
$$
\boxed{G\text{ is solvable, with derived length }2.}
$$
Here <derived length> counts the number of steps needed for the <derived series> to reach the trivial <group>.
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