Calculating $\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}$

I can't figure out how to calculate this limit (or prove it does not exist)

$$ \lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4} $$

I've tried with restrictions on $y=mx$ and curves of the form $y=x^n$. The limit should not exist but even with polar coordinates I can't figure it out

Using polar: $\lvert\dfrac{r^3\cos^2\theta\sin\theta}{r^2(\cos^2\theta+r^2\sin^2\theta)}\rvert=\lvert\dfrac{r\cos^2\theta \sin\theta}{\cos^2\theta +r^2\sin^2\theta}\rvert\le\lvert\dfrac {r\cos^2\theta \sin\theta}{\cos^2\theta}\rvert=\lvert r\sin\theta\rvert\to0$, if $\theta\neq\dfrac{k\pi}2$. But it's easy to see the limit is $0$ when $\theta =\dfrac {k\pi}2$.

If $(x,y) \neq (0,0)$, then we have \begin{align} \left| \dfrac{x^2y}{x^2 + y^4} \right| &= \left| \dfrac{x^2}{x^2 + y^4} \right| \cdot |y| \\ &\leq 1 \cdot |y| \\ &= |y| \end{align} From here it's easy to give an $\varepsilon$-$\delta$ argument for why the limit is $0$.

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 $$ \lim_{(x,y)\to (0,0)}f(x,y)=0 \qquad\qquad\qquad\qquad\qquad\;\;\; $$