$\Leftrightarrow x^{4}+y^{4}-xy^{3}-x^{3}y-(x^{2}+y^{2})(3x-3y)=2(x^{3}-y^{3})-6(x^{2}+y^{2})$
$\Leftrightarrow (x-y)(x^{3}-y^{3})-2(x^{3}-y^{3})-(x^{2}+y^{2})(3x-3y)+6(x^{2}+y^{2})=0$
$\Leftrightarrow (x^{3}-y^{3})(x-y-2)-3(x^{2}+y^{2})(x-y-2)=0$
$\Leftrightarrow (x-y-2)(x^{3}-y^{3}-3x^{2}-3y^{2})=0$
$\Leftrightarrow x^{3}-y^{3}-3x^{2}-3y^{2}=0$
$\Leftrightarrow (x-y)(x^{2}+xy+y^{2})-2(x^{2}+xy+y^{2})-x^{2}-y^{2}+2xy=0$
$\Leftrightarrow (x-y-2)(x^{2}+y^{2}+xy)=(x-y)^{2}$
Ta có $x^{2}+y^{2}+xy=(x+\frac{y}{2})^{2}+\frac{3y^{2}}{4}\geq 0$ mà x-y-2<0
Suy ra (x-y-2)(x2+xy+y2)$\leq 0$
Mặt khác (x-y)2$\geq 0$
Dấu = xảy ra $\Leftrightarrow \left\{\begin{matrix} (x+\frac{y}{2})^{2}=0 & \\ \frac{3y^{2}}{4}=0 & \\ (x-y)^{2}=0& \end{matrix}\right.$
$\Leftrightarrow x=y=0$
Vậy.............