<s>[INST] How to prove $\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$? To prove that $$\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$$
I start with 
$$\left| {\frac{{{x^3}{y^2}}}{{{x^4} + 3{y^4}}}} \right| \leqslant \left| {\frac{{{x^3}{y^2}}}{{{x^4}}}} \right| = \left| {\frac{{{y^2}}}{x}} \right|.$$
But I do not know how to show $| {\frac{{{y^2}}}{x}}|$ is bounded using the hypothesis that $0<|x|<\delta$, $0<|y|<\delta$ and $0<\sqrt{x^2+y^2}<\delta$ since the quarters powers of $x$ and $y$  are very difficult to manage. Even setting $\delta=1$ gets me nowhere.
 [/INST] Another approach is to use the fact that for $x\gt0$, we have $\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0\implies x+\frac1x\ge2$:
$$
\begin{align}
\left|\frac{x^3y^2}{x^4+3y^4}\right|
&=\frac{|x|}{\sqrt3}\frac{x^2(\sqrt[4]3\,y)^2}{x^4+3y^4}\\[9pt]
&=\frac{|x|}{\sqrt3}\frac1{\left(\raise{2pt}{\frac{x}{\sqrt[4]3\,y}}\right)^2+\left(\frac{\sqrt[4]3\,y}{x}\right)^2}\\
&\le\frac{|x|}{2\sqrt3}
\end{align}
$$
</s>