Answer
$\frac{y(x+y)}{(x+1)(x+1)}$.
Work Step by Step
The given expression is
$=\frac{xy-y^2}{x^2+2x+1}\div \frac{2x^2+xy-3y^2}{2x^2+5xy+3y^2}$
Replace divide sign with multiplication sign and reciprocal the right hand side fraction as shown below.
$=\frac{xy-y^2}{x^2+2x+1}\times \frac{2x^2+5xy+3y^2}{2x^2+xy-3y^2}$
Factor each denominator and numerator.
$=xy-y^2$
Factor out common term
$=y(x-y)$
$=x^2+2x+1$
Rewrite middle term.
$=x^2+x+x+1$
Group terms.
$=(x^2+x)+(x+1)$
Factor from each group.
$=x(x+1)+1(x+1)$
Factor out common term.
$=(x+1)(x+1)$
$=2x^2+5xy+3y^2$
Rewrite middle term.
$=2x^2+3xy+2xy+3y^2$
Group terms.
$=(2x^2+3xy)+(2xy+3y^2)$
Factor from each group.
$=x(2x+3y)+y(2x+3y)$
Factor out common term.
$=(2x+3y)(x+y)$
$=2x^2+xy-3y^2$
Rewrite middle term.
$=2x^2+3xy-2xy-3y^2$
Group terms.
$=(2x^2+3xy)+(-2xy-3y^2)$
Factor from each group.
$=x(2x+3y)-y(2x+3y)$
Factor out common term.
$=(2x+3y)(x-y)$
Substitute all factors into the given expression.
$=\frac{y(x-y)}{(x+1)(x+1)}\times \frac{(2x+3y)(x+y)}{(2x+3y)(x-y)}$
Cancel common factors.
The remaining fraction is
$=\frac{y(x+y)}{(x+1)(x+1)}$.
