Different approaches in evaluating the limit $\frac{(x^3+y^3)}{(x^2-y^2)}$ when $(x,y)\to(0,0)$. Note that this question has been previously asked here. I understood the solutions available there but I have two different approaches to this problem, I'm not sure whether they are correct.
I need to know whether both of these solutions are correct and complete. If not, why are they incorrect?
Approach 1

*

*Take path 1 as $y=3x$, hence limit goes to $0$.

*Take path 2 as $y=(-x^3+x^2-y^2)^{1/3}$, hence limit goes to $1$.

Therefore, the limit does not exist.
Is the second path a valid path cause $y$ is not necessarily $0$ when $x=0$?
Approach 2
Take $x=r\cos\theta$ and $y=r\sin\theta$.
We have  $r\frac{\cos^3\theta+\sin^3\theta}{\cos^2\theta-\sin^2\theta}$.
Take path $r = \cos^2\theta-\sin^2\theta$ hence limit goes to $\cos^3\theta+\sin^3\theta$ which is different for every $\theta$ and hence limit cannot exist.
Is this choice of $r$ allowed?
 A: The first one is valid because when $x\to 0$ also $y \to 0$ (see the plot).
The second one is valid also because $r = \cos^2\theta-\sin^2\theta \to 0$ when $\theta \to \frac \pi 4+k\frac \pi 2$ with the limitation that $\cos^2\theta>\sin^2\theta$ (see the plot).
To avoid these kind of check we can use simpler parametric path as for example $x=t$ and $y=t-t^2$ to obtain
$$\frac{x^3+y^3}{x^2-y^2}=\frac{2t^3}{2t^3-t^4}\to 2$$
A: I think one can justify the first approach : you wish to define $y(x)$ so that it satisfies $y^2+y^3 = x^2-x^3$. This is a cubic polynomial in $y$, for a fixed value of $x$, so definitely for each $x$, $y(x)$ can be defined as a real root of this. The problem is : this could have more than one real solution ,  so you need to chose your root $y$ wisely. For example, $y^3+y^2 = 0$ has two roots $-1$ and $0$, and $y^3+y^2 = 0.01$ has three real roots , two very close to $0$ and one close to $-1$.
So, if you define for $x$ in a neighbourhood of $0$, $y(x)$ to be the real root of $y^2+y^3 = x^2-x^3$ closest to $0$, then you need to show that $y(x) \to 0$ as $x \to 0$ . This can be done through some root monotonicity(by which I mean : you show that $|y(x)|$ is decreasing in $|x|$), or some statement of the kind "roots of a polynomial vary continuously with the coefficients" suitably modified. This is proved using the inverse(implicit) function theorem.
I don't show the details, but you can see that with this long winded explanation, we can justify approach $1$, which validates the second path and shows non-existence of the limit.

As for the second approach, I get the point : essentially, you are taking $x = \cos^3 \theta - \cos \theta \sin^2 \theta$ and $y = \cos^2\theta \sin \theta - \sin^3 \theta$ (once I put in the value for $r$). But "limit goes to" does not make sense : where are you moving $\theta$ to? Remember, you can't use any value of $\theta$ because you need to ensure that $(x,y) \to (0,0)$ happens , otherwise the path is not going where you need it to go : one sees that $\theta \to \frac \pi 4$ works. Then your explanation works : the result is $\cos^3 \frac{\pi}{4} + \sin^3 \frac{\pi}{4}$ which is not zero, so you are done.
