What paths to choose to prove that $\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$ does not exist? It's a quite simple question. But I couldn't see it...
How to prove that
$$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$$
doesn't exist?
It's sufficient to show that for different paths through the origin this limit has diferent values (or doesn't exist in some of them). But, what paths to choose?!
 A: If we consider the path: $y^2=x^2+\left |\frac{x^3}{n-x}\right|$ for any real $n \neq 0$ we have :
$$\frac{xy^2}{x^2-y^2}=\frac{x(x^2+\frac{x^3}{n-x})}{\frac{x^2}{n-x}}=\frac{x^2(n-x)+x^3}{x^2}=n $$ 
and here we are done because our pass through to $(0,0)$
A: Note: The original (and accepted) version of this answer was incorrect, owing to a slip in mental algebra on my part.
Try $y=x\sqrt{x+1}$ and $y=0$.
A: Consider $(r,\theta)$, such that $x = r\cos(\theta), y =r\sin(\theta)$.
Then 
$$L(x,y) = \lim_{(x,y)\to 0}\frac{xy^2}{x^2 - y^2} = \lim_{r \to 0} \frac{r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta) -\sin^2(\theta))} = \lim_{r \to 0} r\cdot \frac{\cos(\theta)\sin^2(\theta)}{\cos(2\theta)} = L(r,\theta)$$
So clearly, even though $r\to 0$, the limit is dependant on $\theta$ since $L(r,\theta)$ is undefined for $\theta = \frac\pi4 + \frac{n\pi}2, \ n\in\mathbb{Z}$
A: Generally, you can see that a limit does not exists by choosing differents paths approaching to $(0,0)$. 
Intuitively, imagine your $f(x,y)$ plotted. Then, those paths are functions of one variable, let's say $y=x^2$. Imagine that function plotted on the plane. then, of you do $f(x,x^2)$, you will have the curve $y=x^2$ "projected" on your $f(x,y)$. 
If you approach to $(0,0)$ by different paths (for example, $y=x$, or more generally $y=mx$ (you'll have to look respectively $f(x,x)$ and $f(x,mx)$)) and you find out that the limit approaching $(0,0)$ by those paths are different, then you can say that the limit doesn't exist.
Note that you can't say that a limit exists by looking different paths, even if you've looked $100$ and you seen that the limit is equal in all the cases.
As Brian said, you can choose $y=\frac12\sqrt{x}$ and $y=0$, for example, but those are not the only paths you can choose.
There's another option, the option that used jameselmore. It consists on converting your cartesian coordinates to polar coordinates. If you see that the limit depends on the angle $\theta$, then your limit doesn't exists.
A: In this case just note that the function does not exist anywhere on the line $x=y$, so there cannot be a limit. Following either axis the value is $0$
