Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}} \cr
& {\text{Rewrite the limit using polar coordinates}} \cr
& x = r\cos \theta ,{\text{ }}y = r\sin \theta \cr
& \left( {x,y} \right) \to \left( {0,0} \right),{\text{ so }}r \to 0 \cr
& {\text{Substituting}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}} \Rightarrow \mathop {\lim }\limits_{r \to 0} \frac{{{{\left( {r\cos \theta } \right)}^3} + {{\left( {r\sin \theta } \right)}^3}}}{{{{\left( {r\cos \theta } \right)}^2} + {{\left( {r\sin \theta } \right)}^2}}} \cr
& = \mathop {\lim }\limits_{r \to 0} \frac{{{r^3}{{\cos }^3}\theta + {r^3}{{\sin }^3}\theta }}{{{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}} \cr
& = \mathop {\lim }\limits_{r \to 0} \frac{{r\left( {{{\cos }^3}\theta + {{\sin }^3}\theta } \right)}}{{\left( 1 \right)}} \cr
& = \mathop {\lim }\limits_{r \to 0} r\left( {{{\cos }^3}\theta + {{\sin }^3}\theta } \right) \cr
& {\text{Evaluate the limit when }}r \to 0 \cr
& = \left( 0 \right)\left( {{{\cos }^3}\theta + {{\sin }^3}\theta } \right) \cr
& = 0 \cr
& {\text{Then, we can conclude that}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}} = 0 \cr} $$
