Finding limit of $\frac{x^2y^2}{x^3+y^3}$ as $(x,y)\to (0,0)$ Here, if we go along $x=-y$ then the function itself is not defined, and limit seems to exist for all other paths of approach
Putting in another way, if we use polar coords, limit is:
$$\lim_{r\to 0} \frac{r^4 \sin^2\theta\cos^2\theta}{r^3(\sin^3\theta + \cos^3\theta)}$$
which is zero only when $(\sin^3\theta + \cos^3\theta)$ is nonzero, and function itself is not defined for values where $(\sin^3\theta + \cos^3\theta) = 0$
So here, what do we say? Does the limit exist? or does it not exist
(Additional Question)
Suppose that limit is being calculated $(x,y) \to (0,0)$. Then can we choose one of the paths as, say $x=0$, although in limit we only let variables to "tend" to a value? Is it allowed to use this path?

We have that


*

*$x=y=t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4}{2t^3}=\frac t{2}\to 0$

*$x=t\to 0 \quad y=-t+t^2\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4-2t^5+t^6}{3t^4-3t^5+t^6}=\frac{1-2t+t^2}{3-3t+t^2}\to \frac13$
therefore the limit doesn't exist.
