Answer
$(a)$ limit does not exist
$(b)=0$
\[
(c)=\frac{1}{2}
\]
$(d)=\operatorname{limit} \operatorname{does}$ not exist $\operatorname{as}(x, y) \rightarrow(0,0)$
Work Step by Step
We are given that
\[
f(x, y)=\frac{y x^{2} }{y^{2}+x^{4}}
\]
(a) From the given graph of the function, we can say that limit does not exist at $(x, y) \rightarrow(0,0)$ because there seem to be points near with $z=0$ and other points near (0,0) with $z=\frac{1}{2}$
(b) We have to find the $\lim f(x, y)$ as $(x, y) \rightarrow(0,0)$ along the line $m x=y$
\[
\begin{array}{l}
=\lim _{(x, y) \rightarrow(0,0)} f(x, y) \\
=\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y}{y^{2}+x^{4}} \quad \text { substituting } y=m x \\
=\lim _{x \rightarrow 0} \frac{x^{2}(m x)}{x^{4}+(m x)^{2}} \\
=\lim _{x \rightarrow 0} \frac{m x}{x^{2}+m^{2}} \\
=\frac{m *(0)}{0^{2}+m^{2}} \\
=0
\end{array}
\]
(c) Similarly, finding the limit along $x^{2}=y$
\[
\begin{array}{l}
=\lim _{(x, y) \rightarrow(0,0)} f(x, y) \\
=\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y}{x^{4}+y^{2}} \quad \text { substituting }x^{2}= y \\
=\lim _{x \rightarrow 0} \frac{x^{2}\left(x^{2}\right)}{x^{4}+\left(x^{2}\right)^{2}} \\
=\lim _{x \rightarrow 0} \frac{1}{2} \\
=\frac{1}{2}
\end{array}
\]
(d) Limit must be unique if it exists as we seen in part $b$ and $c$, so $f(x, y)$ can not have a limit as $(x, y) \rightarrow(0,0)$
