Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
   \lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= r^2 \sin^2(t)$, but these attempts were unsatisfactory.
I also have reviewed the similar questions and their answers but there are difference between those functions and mine one, I think the principal difference is that the powers of the denominators are odd.

Actually, that function isn't even bounded near $(0,0)$ and therefore, the limit at that point doesn't exist. You can check that$$\frac{8x^2(-x^3+x^6)^3}{x^9+(-x^3+x^6)^3}=\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}$$and that$$\lim_{x\to0}\left|\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}\right|=\infty.$$
