Answer
Limit does not exist
Work Step by Step
(a) We are given that
\[
f(x, y)=\frac{x^{3} y}{y^{2}+2 x^{6}}
\]
We have to find the $\lim f(x, y)$ as $(x, y) \rightarrow(0,0)$ along the line $m x =y$
\[
\begin{array}{l}
=\lim _{(x, y) \rightarrow(0,0)} f(x, y) \\
=\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3} y}{2 x^{6}+y^{2}} \quad \text { substituting } y=m x \\
=\lim _{x \rightarrow 0} \frac{m x^{2}}{2 x^{4}+m^{2}} \\
=\lim _{x \rightarrow 0} \frac{m(0)^{2}}{(0)^{2}+m^{2}} \\
=0
\end{array}
\]
(b) Similarly, we calculate the limit along $x^{3=y}$
\[
\begin{array}{l}
=\lim _{(x, y) \rightarrow(0,0)} f(x, y) \\
=\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3} y}{2 x^{6}+y^{2}} \quad \text { substituting } x^{3}=y \\
=\lim _{x \rightarrow 0} \frac{x^{3} * x^{3}}{2 x^{6}+x^{6}} \\
=\lim _{x \rightarrow 0} \frac{1}{3} \\
=\frac{1}{3}
\end{array}
\]
Since along different curves passing through (0,0) we get different limits, the limit of the function does not exist.
