Answer
$ - \frac{1}{2}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^3} - {x^3}{y^2}}}{{{x^2} - {y^2}}}} \right) \cr
& {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr
& {\text{Substitute 1 for }}x{\text{ and }}1{\text{ for }}y \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^3} - {x^3}{y^2}}}{{{x^2} - {y^2}}}} \right) = \frac{{{{\left( 1 \right)}^2}{{\left( 1 \right)}^3} - {{\left( 1 \right)}^3}{{\left( 1 \right)}^2}}}{{{{\left( 1 \right)}^2} - {{\left( 1 \right)}^2}}} = \frac{0}{0} \cr
& {\text{Factoring the numerator and denominator of }}f\left( {x,y} \right) \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^3} - {x^3}{y^2}}}{{{x^2} - {y^2}}}} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^2}\left( {y - x} \right)}}{{\left( {x + y} \right)\left( {x - y} \right)}}} \right) \cr
& = - \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^2}\left( {y - x} \right)}}{{\left( {x + y} \right)\left( {y - x} \right)}}} \right) \cr
& {\text{Cancel the common factor }}y - x \cr
& = - \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^2}}}{{x + y}}} \right) \cr
& {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr
& = - \frac{{{{\left( 1 \right)}^2}{{\left( 1 \right)}^2}}}{{\left( 1 \right) + 1}} \cr
& = - \frac{1}{2} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,1} \right)} \left( {\frac{{{x^2}{y^3} - {x^3}{y^2}}}{{{x^2} - {y^2}}}} \right) = - \frac{1}{2} \cr} $$
