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

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

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 $$ \lim_{(x,y)\to(0,0)}\frac{x^3y}{x^2+y^4}=0. $$