Method to prove limit in $\mathbb{R}^2$ Given is the following limit $\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$.
Now, it holds that
$0\leq\left\vert\frac{x^3y}{x^4+y^2}\right\vert\leq \frac{x^3y}{y^2}=\left\vert\frac{x^3}{y}\right\vert$.
It is clear that $(x,y)\mapsto\frac{x^3}{y}$ is homogeneous on $\mathbb{R^2}\setminus\{\vec0\}$. This implies that $\lim_{(x,y)\to(0,0)}\frac{x^3}{y}=0$.
The squeeze lemma would now imply that
$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}=0$.
Where does it go wrong?

Let $S=\frac {|x^3y|}{x^4+y^2}.$ We have $x^2|y|\leq \frac {(x^2+|y|)^2}{4}$ so $$S=\frac {|x|\cdot x^2|y|}{x^4+y^2} \leq \frac {|x|(x^2+|y|)^2}{4(x^4+y^2)}=$$ $$=\frac {|x|(x^4+y^2)+2|x^3y|}{4(x^4+y^2)}=$$ $$ =\frac {|x|}{4}+\frac {S}{2}.$$  $$ \text {Therefore }\quad S\leq \frac {|x|}{2}.$$ 
Not as slick as the answer from Dr.MV but it works. 
