Answer
$ - 2$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \frac{{{x^2}y + x{y^3}}}{{{x^2} - {y^2}}} \cr
& {\text{ }}f\left( {x,y} \right) = \frac{{{x^2}y + x{y^3}}}{{{x^2} - {y^2}}}\cr
&{\text{ is a rational function}}\cr
&{\text{and }}f\left( {2, - 1} \right) = 3 \cr
& {\text{The denominator is not 0,}}\cr
&{\text{therefore we can find the limit by direct }} \cr
& {\text{substitution}}{\text{.}} \cr
& {\text{Substitute 2 for }}x{\text{ and}} - 1{\text{ for }}y \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \frac{{{x^2}y + x{y^3}}}{{{x^2} - {y^2}}} = \frac{{{{\left( 2 \right)}^2}\left( { - 1} \right) + \left( 2 \right){{\left( { - 1} \right)}^3}}}{{{{\left( 2 \right)}^2} - {{\left( { - 1} \right)}^2}}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \frac{{{x^2}y + x{y^3}}}{{{x^2} - {y^2}}} = \frac{{ - 6}}{3} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \frac{{{x^2}y + x{y^3}}}{{{x^2} - {y^2}}} = - 2 \cr} $$
