Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} x\root 3 \of {{y^3} + 2x} \cr
& {\text{Using the limit laws }} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} x\root 3 \of {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} \left( {{y^3} + 2x} \right)} \cr
& = \left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} x} \right)\left( {\root 3 \of {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} {y^3} + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4, - 2} \right)} 2x} } \right) \cr
& {\text{then}}{\text{, substituting }}4{\text{ for }}x{\text{ and }} - 2{\text{ for }}y \cr
& = \left( 4 \right)\left( {\root 3 \of {{{\left( { - 2} \right)}^3} + 2\left( 4 \right)} } \right) \cr
& = \left( 4 \right)\left( {\root 3 \of { - 8 + 8} } \right) \cr
& = 0 \cr} $$
