Something is wrong with this proof, limit $\lim\limits_{(x,y) \to (0,0)} \frac{xy^3}{x^4 + 3y^4}$ Could someone please tell me what is wrong with this proof?

Show that $\lim\limits_{(x,y) \to (0,0)}  \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.

I know it is wrong because the limit doesn't exist, but this proof is contradicting me
Proof
Case 1
Assume for $x,y > 0$, then $x^4 + 3y^4 > x^4 > x > 0$
$$\begin{align*}
0 < x < x^4 + 3y^4  &\iff 0 < \dfrac{x}{x^4 + 3y^4} < 1 \\
& \iff 0 < \dfrac{x|y^3|}{x^4 + 3y^4} < |y^3| \\
& \iff \lim\limits_{(x,y) \to (0,0)} 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} <  \lim\limits_{(x,y) \to (0,0)}|y^3|\\
&\iff 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < 0
\end{align*}$$
Case 2. WLOG Assume $x,y <0$ and combine both cases.
What's wrong the ppoof? I don't find the flaw

The easiest way to see that a limit doesn't exist is to write
$$
\frac{xy^3}{x^4+y^4}=\frac{x/y}{(x/y)^4+1}\tag{1}
$$
Using $(1)$, it is easy to see that, over any circle around the origin, $\frac{xy^3}{x^4+y^4}$ can take on any value that $\frac{t}{t^4+1}$ can; that is, any value in $\left[{-}\sqrt[4]{\frac{27}{256}}\;,\;\sqrt[4]{\frac{27}{256}}\right]$. Therefore,
$$
\lim_{x,y\to0}\frac{xy^3}{x^4+y^4}\tag{2}
$$
does not exist.
