<s>[INST] Show this function of two variables $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at (0,0) Is the function $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$  continuous at $(0, 0)$?
Trying to use the epsilon-delta definition of continuity to prove this but can't figure it out.
$\sqrt{x^2+y^2} < \delta$ implies $\left|\frac{2x^2y^3}{x^4+y^6}\right| < \epsilon$.
Using polar coordinates $r < \delta$ implies 
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| < \epsilon^4$$
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| \leq \left|\frac{2r}{\cos^4(\theta)+r^2\sin^6(\theta)}\right|$$
But from here I cant simplify the equation further.
 [/INST] **Hint:**As you approach $(0,0)$ along $y^3=mx^2$, you get $lim_{(x,y)→(0,0)}f(x,y)=2m/(1+m^2)$ which depends on arbitrary $m$.
</s>