Set $a=xy^2$ and $b=a^x = y^2x$, so that $ab= xyx$ and $ba =y^2 x^2 y^2$. But $xyxyxy = 1$ so $xyx = y^{-1} x^{-1} y^{-1} = y^2 x^2 y^2$, so $a$ and $b$ commute, and so form a normal abelian subgroup $A$ that is a quotient of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, and which together with either $x$ or $y$ generates $G$.
Check that the semi direct product of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ with $x(a) = b$, $x(b)= (ab)^{-1}$ satisfies the relations so that $G$ is a generalization of the alternating group of order 12, having in general order $3n^2$ instead of $3\cdot 2^2$.
If you omit the last relation (set $n=0$), then you get the following faithful integral matrix representation of the group. To include the last relation, just interpret the matrices mod $n$ to get a faithful matrix rep over $\mathbb{Z}/n\mathbb{Z}$.
$ x = \left[\begin{smallmatrix} 0 & 1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right] \quad y = \left[\begin{smallmatrix} 0 & 1 & -1 \\-1 & -1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right] \quad a = \left[\begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right] \quad b = \left[\begin{smallmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix}\right] $