Limit of $\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y^2}{x^3+y^3}$

I want to know whether $\lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^3+y^3}$ exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example,

$f(x,mx^2) = \dfrac{m^2x^3}{1+m^3x^3}$

But that doesn't show that the limit is 0.

Hint: There is real trouble if $x$ is the negative of $y$. If you don't want to take the easy way out and note that the function is not defined when $x=-y\ne 0$, let $(x,y)$ approach $(0,0)$ along the parametric path $x=t+t^2$, $y=-t+t^2$. If you want more dramatic behaviour, use the path $(t+t^3, -t+t^3)$.

Hint

Choose any $\alpha\ne0$. As $t\to-1$, consider the path $$ (x,y)=\alpha(t^3+1)\left(\frac1{t^2},\frac1{t}\right)\to(0,0) $$ For $\alpha=0$ use $$ (x,y)=(t^3+1,0)\to(0,0) $$ More Hint

For all $t\ne-1$, $\dfrac{x^2y^2}{x^3+y^3}=\alpha$.

Hint: Use $\sqrt[3]{\frac{x^3+y^3}2}\ge \sqrt{xy}$.