Answer
$${\text{The limit does not exist}}{\text{}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2}y}}{{{x^4} + {y^2}}} \cr
& {\text{Evaluate and substitute 0 for }}x{\text{ and }}0{\text{ for }}y \cr
& = \frac{{{{\left( 0 \right)}^2}\left( 0 \right)}}{{{{\left( 0 \right)}^4} + {{\left( 0 \right)}^2}}} \cr
& = \frac{0}{0} \cr
& {\text{First}},{\text{ we consider the limit along the path }}y = 0.{\text{ We have}} \cr
& \mathop {\lim }\limits_{\left( {x,0} \right) \to \left( {x,0} \right)} \frac{{{x^2}y}}{{{x^4} + {y^2}}} = \frac{{{x^2}\left( 0 \right)}}{{{x^4} + {{\left( 0 \right)}^2}}} = 0 \cr
& {\text{Similarly}},{\text{ for the path }}x = 0,{\text{ we have}} \cr
& \mathop {\lim }\limits_{\left( {0,y} \right) \to \left( {0,y} \right)} \frac{{{x^2}y}}{{{x^4} + {y^2}}} = \frac{{{{\left( 0 \right)}^2}y}}{{{{\left( 0 \right)}^4} + {{\left( y \right)}^2}}} = 0 \cr
& {\text{Just because the limits along the first two paths you try are }} \cr
& {\text{the same does not mean that the limit exists}} \cr
& {\text{We may simply need to look at more paths}}. \cr
& {\text{for the path }}y = x,{\text{ we have}} \cr
& \mathop {\lim }\limits_{\left( {x,x} \right) \to \left( {0,0} \right)} \frac{{{x^2}\left( x \right)}}{{{x^4} + {{\left( x \right)}^2}}} = \mathop {\lim }\limits_{\left( {x,x} \right) \to \left( {0,0} \right)} \frac{{{x^2}}}{{{x^2} + 1}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{{x^2} + 1}} = 0 \cr
& {\text{for the path }}y = {x^2},{\text{ we have}} \cr
& \mathop {\lim }\limits_{\left( {x,{x^2}} \right) \to \left( {0,0} \right)} \frac{{{x^2}\left( {{x^2}} \right)}}{{{x^4} + {{\left( {{x^2}} \right)}^2}}} = \mathop {\lim }\limits_{\left( {x,x} \right) \to \left( {0,0} \right)} \frac{{{x^4}}}{{2{x^4}}} = \frac{1}{2} \cr
& {\text{Since the limit along this path doesn}}{\text{'t match the}} \cr
& {\text{limit along the first paths, the limit does not exist}}{\text{.}} \cr} $$
