Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$.
Let $x^2+y^2=r^2$, with $0 < r \le 1$.
If $x\ne 0$, then
\begin{align*}
|f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt]
&\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt]
&=\left|\frac{y}{1+y^4}\right|\\[4pt]
&\le |y|\\[4pt]
&\le r\\[4pt]
\end{align*}
and if $x=0$, then $y\ne 0$, so
$$
f(x,y)=\frac{0}{y^4}=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$
In either case, we have $|f(x,y)|\le r$.
Letting $r$ approach zero from above, it follows that
$$
\lim_{(x,y)\to (0,0)}f(x,y)=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$