$\Delta=\begin{vmatrix}x& x+y & x+2y\\x+2y & x & x+y\\x+y & x+2y & x\end{vmatrix}$
Step 1: Apply $R_1\rightarrow R_1+R_2+R_3$
$\Delta=\begin{vmatrix}3x+3y& 3x+3y & 3x+3y\\x+2y & x & x+y\\x+y & x+2y & x\end{vmatrix}$
Step 2:Apply $C_2\rightarrow C_2-C_3$ and $C_3\rightarrow C_3-C_1$
$\Delta=3(x+y)\begin{vmatrix}1& 0 & 0\\x+2y & -y & -y\\x+y & 2y & -y\end{vmatrix}$
Therefore $\Delta=3(x+y)(3y^2)$
$\qquad\qquad\;\;=9y^2(x+y)$
Hence proved.