Answer
(a) as $\left( {x,y} \right) \to \left( {0,0} \right)$ along any line $y = mx$, the limit
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3}y}}{{{x^6} + 2{y^2}}}=0$
(b) as $\left( {x,y} \right) \to \left( {0,0} \right)$ along the curve $y = {x^3}$, the limit
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3}y}}{{{x^6} + 2{y^2}}}= \frac{1}{3}$
Since $f\left( {x,y} \right)$ does not approach one limit as $\left( {x,y} \right) \to \left( {0,0} \right)$, the limit does not exist.
Work Step by Step
(a) Evaluate the limit along any line $y = mx$:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3}y}}{{{x^6} + 2{y^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\left( {mx} \right)}}{{{x^6} + 2{{\left( {mx} \right)}^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{m}{{{x^2} + 2\frac{{{m^2}}}{{{x^2}}}}}$
$ = 0$
(b) Evaluate the limit along the curve $y = {x^3}$:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3}y}}{{{x^6} + 2{y^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\left( {{x^3}} \right)}}{{{x^6} + 2{{\left( {{x^3}} \right)}^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + 2}}$
$ = \frac{1}{3}$
Since the limit does not equals $0$, $f\left( {x,y} \right)$ does not approach one limit as $\left( {x,y} \right) \to \left( {0,0} \right)$. Therefore, $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right)$ does not exist.
