Answer
$$ - \frac{{6{x^2} + {y^2}}}{{{x^3}{{\left( {x + y} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& \frac{{\frac{1}{{{{\left( {x + y} \right)}^3}}} - \frac{1}{{{x^3}}}}}{y} \cr
& {\text{Subtract fractions in the numerator}} \cr
& = \frac{{\frac{{{x^3} - {{\left( {x + y} \right)}^3}}}{{{x^3}{{\left( {x + y} \right)}^3}}}}}{y} \cr
& = \frac{{\frac{{{x^3} - \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right)}}{{{x^3}{{\left( {x + y} \right)}^3}}}}}{y} \cr
& = \frac{{\frac{{{x^3} - {x^3} - 3{x^2}y - 3x{y^2} - {y^3}}}{{{x^3}{{\left( {x + y} \right)}^3}}}}}{y} \cr
& = \frac{{\frac{{ - 6{x^2}y - {y^3}}}{{{x^3}{{\left( {x + y} \right)}^3}}}}}{y} \cr
& = \frac{{ - 6{x^2}y - {y^3}}}{{{x^3}y{{\left( {x + y} \right)}^3}}} \cr
& {\text{Simplify}} \cr
& = \frac{{ - 6{x^2} - {y^2}}}{{{x^3}{{\left( {x + y} \right)}^3}}} \cr
& = - \frac{{6{x^2} + {y^2}}}{{{x^3}{{\left( {x + y} \right)}^3}}} \cr} $$
