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