How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$ 
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$

First I tried to contradict by using $y = mx$ , but I found that the limit exists.
Secondly I tried to use polar coordinates, $x =  \cos\theta $ and $y = \sin\theta$,
And failed .. How would you prove this limit equals $0$?
 A: Note that, when $x$ and $y$ are both less than $1$, we have 
$\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| < \frac{1}{2}$
multiplying by $|x|$ we get
$\displaystyle \left|\frac{x^3y}{x^4+y^2}\right| < \frac{1}{2}|x|$
Result follows from squeeze theorem
A: Cauchy inequality:
$$x^4+y^2\ge 2 x^2 |y|$$
Thus
$$
\left|\frac{x^3y}{x^2+y^4}\right|\le \frac{|x|}{2}.
$$
But the RHS tends to $0$ as $(x,y)\to (0,0)$, and hence
$$
\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^2+y^4}=0.
$$
A: Observe that $x^4 + y^2 \geq |x^2y|$  (for instance, because $x^4+y^2+2x^2y = (x^2+y)^2\geq0$ and $x^4+y^2-2x^2y = (x^2-y)^2 \geq0$).  Hence $\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| \leq 1$ when $(x,y)\neq (0,0)$ and thus $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^3y}{x^4+y^2}\right| \leq \lim_{(x,y)\rightarrow(0,0)} |x| = 0,$$ so the limit is 0 by the squeeze theorem.
A: Let $y=kx^2$, then the expression is $\frac{kx}{1+k^2}$.
$|\frac{k}{1+k^2}|\leq 1/2$ so the expression is bounded by $|x/2|$.  The path along $x=0$ can be done separately.
A: With polar coordinates we're cool, too:
$$x=r\cos\theta\;\;,\;\;y=r\sin\theta$$ 
$$\frac{x^3y}{x^4+y^2}=\frac{r^4\cos^3\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\frac{r^2\cos^3\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\xrightarrow[r\to 0]{}\frac0{0+\sin^2\theta}=0$$
But what if $\;\sin^2\theta=0\;$ ? Well, then also $\;\sin\theta=0\;$ and the first expression above's already zero from the beginning...
