Answer
$x^3y^3(xy-1)(x^2y^2+xy+1)$
Work Step by Step
The formula for factoring the sum of two cubes is: $A^3+B^3=(A+B)(A^2-AB+B^2)$.
The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$.
Hence here: $x^6y^6-x^3y^3=\\=x^3y^3(x^3y^3-1)\\=x^3y^3((xy)^3-1^3)\\=x^3y^3(xy-1)((xy)^2+xy+1^2)\\=x^3y^3(xy-1)(x^2y^2+xy+1)$