prove that $f(x,y)= \frac{x^3y}{x^4+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$ is continous at $(0,0)$ How to prove that $f: \mathbb R^2 \to \mathbb R$ $f(x,y)= \frac{x^3y}{x^4+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$ is continous at $(0,0)$?
By definition: Let $\epsilon>0$ I need to prove that $\exists \delta>0$ so that $\forall \vec x\in B_\delta(0,0)$ then $|f(x,y)-0|<\epsilon$
$|f(x,y)|=|{x^3y\over x^4+y^2}|$=${|x^3y|\over x^4+y^2}$ I know that $|x|,|y|\le ||\vec x||$ so $|x^3y|\le ||\vec x||^4$ then ${|x^3y|\over x^4+y^2}\le {||\vec x||^4\over x^4+y^2}$ but I don´t know how to proceed from here
Any ideas? 

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
$\displaystyle \left|\frac{x^3y}{x^4+y^2}\right| < \frac{1}{2}|x|$
Now result follows from squeeze theorem
