Answer
$$ - \frac{{2x + y}}{{{x^2}{{\left( {x + y} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{\frac{1}{{{{\left( {x + y} \right)}^2}}} - \frac{1}{{{x^2}}}}}{y} \cr
& {\text{Subtract fractions in the numerator}} \cr
& = \frac{{\frac{{{x^2} - {{\left( {x + y} \right)}^2}}}{{{x^2}{{\left( {x + y} \right)}^2}}}}}{y} \cr
& = \frac{{\frac{{{x^2} - {x^2} - 2xy - {y^2}}}{{{x^2}{{\left( {x + y} \right)}^2}}}}}{y} \cr
& = \frac{{\frac{{ - 2xy - {y^2}}}{{{x^2}{{\left( {x + y} \right)}^2}}}}}{y} = \frac{{ - 2xy - {y^2}}}{{{x^2}y{{\left( {x + y} \right)}^2}}} \cr
& {\text{Simplify}} \cr
& = \frac{{ - 2x - y}}{{{x^2}{{\left( {x + y} \right)}^2}}} \cr
& = - \frac{{2x + y}}{{{x^2}{{\left( {x + y} \right)}^2}}} \cr} $$
