Why does the limit $\frac{xy}{y-x^3}$ when $(x,y) \rightarrow (0,0)$ equal $0$? After I've tried a lot of paths, always, I've ended up with $0$. 
Accordingly the wolfram alpha, it is correct. 
However, I don't know how to prove it using, for example, the squeezing method.
What is it the best approach to deal with it?
I thank for any help.
 A: There is a reasonable argument for staying away from $y=x^3$. However, let $y=x^3+x^4$. The limit along that path is $1$. You can play with that idea and get various values.
A: Take $x_n = {1 \over n}$ and $y_n = {1 \over n^3-n^2}$.
Note that $x_n \to 0, y_n \to 0$.
Now note that ${x_n y_n \over y_n - x_n^3} = n$.
Note that $\lim_n n = \infty$.
Alternatively, pick some $c$ and look at the path $(x, {cx^3 \over c-x})$, and
note that the limit along this path is $c$ (adjust appropriately to avoid $x=c$).
A: Simple fact: If you have continuous functions $f(x,y),g(x,y)$ in the open first quadrant $U$, with $f$ nonzero in $U$ and $g = 0$ only along a curve $\gamma$ in $U$ that approaches $(0,0),$ then as $(x,y) \to (0,0)$ within $U \setminus \gamma,$
$\lim f(x,y)/g(x,y)$ does not exist. Proof: Take any point on $\gamma.$ Then $f/g$ blows up at that point. It follows that $f/g$ is unbounded in every $D((0,0),r)\cap (U \setminus \gamma),$ so the limit cannot exist. (In the present problem, $f(x,y) = xy, g(x,y) = y-x^3, \gamma = (x,x^3).$)
