Solution:

$\begin{array}{l}
\left|\begin{array}{ccc}
\mathrm{x} & \mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} \\
\mathrm{x}+2 \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right| \\
=\left|\begin{array}{cccc}
3(\mathrm{x}+\mathrm{y}) & 3(\mathrm{x}+\mathrm{y}) & 3(\mathrm{x}+\mathrm{y}) \\
\mathrm{x}+2 \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|\left[\mathrm{R}_{1}^{\prime}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}\right] \\
=3(\mathrm{x}+\mathrm{y})\left|\begin{array}{ccc}
\mathrm{x}+2 \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|\left[\mathrm{R}_{1}^{\prime}=\mathrm{R}_{1} / 3(\mathrm{x}+\mathrm{y})\right] \\
=3(\mathrm{x}+\mathrm{y})\left|\begin{array}{ccc}
\mathrm{y} & -2 \mathrm{y} & \mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|
\end{array}$
$=3 \mathrm{y}(\mathrm{x}+\mathrm{y})\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|\left[\mathrm{R}_{2}^{\prime}=\mathrm{R}_{2} / \mathrm{y}\right]$
$\begin{array}{l}
=3 y(x+y)\left|\begin{array}{ccc}
0 & 3 & 0 \\
1 & -2 & 1 \\
x+y & x+2 y & x
\end{array}\right|\left[R_{1}^{\prime}=R_{1}-R_{2}\right] \\
=3 y(x+y)[0+3(x+y)-x+0][\text { expansion by first row }] \\
=3 y(x+y)(3 y)=9 y^{2}(x+y)
\end{array}$