Answer
$\displaystyle \frac{(2x+y)(3x+y)}{(3x-y)(x+3y)}$
Work Step by Step
Factor what we can:
$ 2x^{2}-3xy-2y^{2}=2x^{2}-4xy+xy-2y^{2}=\qquad$ ... factor in pairs
$=2x(x-2y)+y(x-2y)$
$=(x-2y)(2x+y)$
$ 3x^{2}-2xy-y^{2}=3x^{2}-3xy+xy-y^{2}=\qquad$ ... factor in pairs
$=3x(x-y)+y(x-y)$
$=(x-y)(3x+y)$
$ 3x^{2}-4xy+y^{2}=3x^{2}-3xy-xy+y^{2}\qquad$ ... factor in pairs
$=3x(x-y)-y(x-y)$
$=(x-y)(3x-y)$
$ x^{2}+xy-6y^{2}=x^{2}+3xy-2xy-6y^{2}\qquad$ ... factor in pairs
$=x(x+3y)-2y(x+3y)$
$=(x+3y)(x-2y)$
Rewrite the problem:
$\displaystyle \frac{(x-2y)(2x+y)}{(x-y)(3x-y)}\cdot\frac{(x-y)(3x+y)}{(x+3y)(x-2y)}=\qquad$ ... reduce common factors
$=\displaystyle \frac{(1)(2x+y)}{(1)(3x-y)}\cdot\frac{(1)(3x+y)}{(x+3y)(1)}=\qquad$
= $\displaystyle \frac{(2x+y)(3x+y)}{(3x-y)(x+3y)}$
