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.
 A: Say that $0 < \sqrt{x^2+y^2} < \delta$. We can conclude that $|x|<\delta$ and $|y|<\delta$. 
Now $x^4 + 3y^4 \geq x^4 + y^4 \geq 2x^2y^2\geq x^2y^2 $
Thus, 
$$ \left| \frac{x^3y^2}{x^4 + 3y^4} \right| \leq \frac{|x|^3y^2}{x^2y^2} = |x| < \delta$$
Now given any $\varepsilon > 0$ if we choose $\delta = \varepsilon$ this would verify the definition of limits. 
Note: One is not allowed to work with the function $y^2/x$, as you try to do, because it is not locally defined in a punctured disk.
A: Another approach is to use the fact that for $x\gt0$, we have $\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0\implies x+\frac1x\ge2$:
$$
\begin{align}
\left|\frac{x^3y^2}{x^4+3y^4}\right|
&=\frac{|x|}{\sqrt3}\frac{x^2(\sqrt[4]3\,y)^2}{x^4+3y^4}\\[9pt]
&=\frac{|x|}{\sqrt3}\frac1{\left(\raise{2pt}{\frac{x}{\sqrt[4]3\,y}}\right)^2+\left(\frac{\sqrt[4]3\,y}{x}\right)^2}\\
&\le\frac{|x|}{2\sqrt3}
\end{align}
$$
A: Another idea is using polar coordinates. Then
\begin{align}
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^4+3y^4} &= \lim_{r \to 0^+} \frac{r^5\cos^3x\sin^2x}{r^4\cos^4x+3r^4\sin^4x} \\
&=\lim_{r \to 0^+}r \frac{\cos^3x\sin^2x}{\cos^4x + 3\sin^4x}
\end{align}
Now, since 
$$0 \leq \cos^3x\sin^2x \leq 1$$
and
$$\frac{3}{4} \leq \cos^4x+3\sin^4x \leq 3$$
We have
$$\lim_{r \to 0^+}0 \cdot r \leq \lim_{r \to 0^+}r \frac{\cos^3x\sin^2x}{\cos^4x + 3\sin^4x} \leq \lim_{r \to 0^+} \frac{4r}{3}$$
