Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \ln \left( {1 + {x^2}{y^3}} \right) \cr
& {\text{Using the limit laws }} \cr
& = \ln \left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left[ {1 + {x^2}{y^3}} \right]} \right) \cr
& = \ln \left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} 1 + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} {x^2}{y^3}} \right) \cr
& {\text{then}}{\text{, substituting }}0{\text{ for }}x{\text{ and }}0{\text{ for }}y \cr
& = \ln \left( {1 + {{\left( 0 \right)}^2}{{\left( 0 \right)}^3}} \right) \cr
& {\text{simplifying}} \cr
& = \ln \left( 1 \right) \cr
& = 0 \cr} $$
