Система уравнений с двумя
неизвестными
|x^2y-xy^2=6
|x^3-y^3=-9

$left { {{x^2y-xy^2=6} atop {x^3-y^3=-9}} right.$

Выносим общий множитель
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$left { {{xy(x-y)} atop {(x-y)(x^2+xy+y^2)=-9}} right. to left { {{-3xy(x-y)-2(x^2+xy+y^2)(x-y)=-3cdot 6-2(-9)} atop {(x^2+xy+y^2)(x-y)=-9}} right.$

$left { {{3xy+2(x^2+xy+y^2)=0} atop {x^3-y^3=-9}} right. to left { {{2x^2+5xy+2y^2=0} atop {x^3-y^3=-9}} right. to left { {{2x^2+xy+4xy+2y^2=0} atop {x^3-y^3=-9}}right. to to left { {{(2x+y)(x+2y)=0} atop {x^3-y^3=-9}} right.$
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Имеем 2 системы

$left { {{2x+y=0} atop {x^3-y^3=-9}} right.$ и $left { {{x+2y=0} atop {x^3-y^3=-9}} right.$

Случай 1.
$left { {{2x+y=0} atop {x^3-y^3=-9}} right. to left { {{y=-2x} atop {x^3-(-2x)^3=-9}} right. x^3+8x^3=-9 9x^3=-9 x^3=-1 x_1=-1 y_1=-2cdot (-1)=2$

случай 2.
$left { {{x+2y=0} atop {x^3-y^3=-9}} right. to left { {{x=-2y} atop {(-2y)^3-y^3=-9}} right. -8y^3-y^3=-9 -9y^3=-9 y_2=1 x_2=-2$

ответ: $(-1;2),,,,,(-2;1).$