Using the $\epsilon - \delta$ definition for limit show that $\lim_{(x,y)\to (0,0)} \frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}=0$. Using the $\epsilon - \delta$ definition for limit show that $\lim_{(x,y)\to (0,0)} \frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}=0$.
My answer: $\frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2} \leq \frac{3x^3+2y^3}{x^2+y^2} \leq 3\frac{x^3+y^3}{x^2+y^2}$.
I somehow need need to relate $x^3+y^3$ to $x^2+y^2$ so that I can bound the fraction. But I cant seem to arrive at a relation that would work. Help please!

Suppose that $(x,y)\ne (0,0)$. Then 
$$\small \left|\frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}\right|\le \left|\frac{3x^3+2y^3}{x^2+y^2}\right|\le 3|x|\frac{x^2}{x^2+y^2}+2|y|\frac{y^2}{x^2+y^2}\le 3|x|+2|y|\le 5\sqrt{x^2+y^2}.$$
Now finding a suitable $\delta$ will be straightforward.
