jmc

From the given equation, $x^3{y}^3-x^3-y^3=3x^2{y}^2,$ or $$x^3{y}^3-x^3-y^3-3x^2{y}^2=0.$$We have the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$Taking $a=xy,$ $b=-x,$ and $c=-y,$ we get $$x^3{y}^3-x^3-y^3-3x^2{y}^2=(xy-x-y)(a^2+b^2+c^2-ab-ac-bc)=0.$$If $xy-x-y=0,$ then $$(x-1)(y-1)=xy-x-y+1=1.$$If $a^2+b^2+c^2-ab-ac-bc=0,$ then $2a^2+2b^2+2c^2-2ab-2ac-2bc=0,$ which we can write as $$(a-b{)}^2+(a-c{)}^2+(b-c{)}^2=0.$$This forces $a=b=c,$ so $xy=-x=-y.$ We get that $x=y,$ so $x^2+x=x(x+1)=0.$ Hence, $x=0$ or $x=-1.$ From the given condition, we cannot have $x=0,$ so $x=-1,$ and $y=-1,$ so $(x-1)(y-1)=4.$

Thus, the possible values of $(x-1)(y-1)$ are 1 and 4, and their sum is $\boxed{5}.$