Ta có:
$x^2+2y^2+3xy-x-y+3=0$
$\Leftrightarrow (x^2+xy)+(2y^2+2xy)-(x+y)=-3$
$\Leftrightarrow x(x+y)+2y(x+y)-(x+y)=-3$
$\Leftrightarrow (x+y)(x+2y-1)=-3$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x+y=1\\x+2y-1=-3\end{array}\right.\\\left\{\begin{array}{l}x+y=-1\\x+2y-1=3\end{array}\right.\\\left\{\begin{array}{l}x+y=3\\x+2y-1=-1\end{array}\right.\\\left\{\begin{array}{l}x+y=-3\\x+2y-1=1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=4\\y=-3\end{array}\right.\\\left\{\begin{array}{l}x=-6\\y=5\end{array}\right.\\\left\{\begin{array}{l}x=6\\y=-3\end{array}\right.\\\left\{\begin{array}{l}x=-8\\y=5\end{array}\right.\end{array}\right.$