Answer
$x^{2}+y^{2}+2xy-2x^{2}y-2xy^{2}+x^{2}y^{2}$
Work Step by Step
We can rewrite this as a special formula (square of a difference)
$[(x+y)-xy]^{2}=...$
$(I-II)^{2}= I^{2}-2\cdot I\cdot II+II^{2}$,
where $I=(x+y),\quad II=xy$
$I^{2}=(x+y)^{2}$=special f.$=x^{2}+2xy+y^{2}$
$2\cdot I\cdot II=2(x+y)xy=2xy(x+y)=2x^{2}y+2xy^{2}$
$II^{2}=(xy)^{2}=x^{2}y^{2}$
$...=x^{2}+2xy+y^{2}-(2x^{2}y+2xy^{2})+x^{2}y^{2}$
$=x^{2}+2xy+y^{2}-2x^{2}y-2xy^{2}+x^{2}y^{2}$
... look for like terms to add ... none