= Solution
Let $F=\mathbb Ze_1\oplus\mathbb Ze_2$ and $N=\mathbb Z(6e_1+9e_2)$. The precise presented <group> is $A=F/N$, with $x=e_1+N$ and $y=e_2+N$. It is abelian and has exactly the imposed additive relation and its integer multiples.
Change the <integral> <basis> to $u=2e_1+3e_2$, $v=e_1+2e_2$. The change-of-basis <determinant> is $2\cdot2-3\cdot1=1$, so these are a <basis> of the <free abelian group>. Explicitly, $e_1=2u-3v$ and $e_2=-u+2v$. The relation generator becomes $6e_1+9e_2=3u$. Therefore
$$
\boxed{A\cong(\mathbb Zu/3\mathbb Zu)\oplus\mathbb Zv\cong\mathbb Z/3\mathbb Z\oplus\mathbb Z.}
$$
The image of $u$ is nonzero, since $(2,3)$ is not an integer multiple of $(6,9)$, but three times it vanishes. Thus $A$ has an element of exact order three. A <free abelian group> has no nonzero finite-order element, so \b[$A$ is not free abelian].
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