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 statement $x^4>x$ is clearly wrong especially when $x$ is in the neighborhood of $0$. 
Also, you need to look at cases $x>0,y<0$ and $x<0,y>0$. But the main error is the statement $x^4 > x$.
EDIT
The answer is that the function is not continuous at the origin. This can be seen as follows. Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze.
Approach the origin along the straight line $y=mx$. Then we have that $$\lim_{\overset{x \rightarrow 0}{y \rightarrow 0}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{\overset{x \rightarrow 0}{y = mx}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{x \rightarrow 0} \dfrac{x(mx)^3}{x^4 + 3(mx)^4} \\ = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{x^4 + 3m^4 x^4} = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{(1 + 3m^4) x^4} = \lim_{x \rightarrow 0} \dfrac{m^3}{(1 + 3m^4)} = \dfrac{m^3}{1+3m^4}$$
Hence, approaching the origin along different straight lines give different answers.
For instance, if we approach the origin along the line $y = x$, the limit is $\dfrac14$.
If we approach the origin along the line $y = -x$, the limit is $-\dfrac14$.
If we approach the origin along the line $y = 0$, the limit is $0$.
This proves that the function is not continuous at the origin.
