Two paths that show that $\frac{x-y}{x^2 + y^2}$ has no limit when $(x,y) \rightarrow (0,0)$

I'm having a difficult time trying to find two different paths that give me different limits for the following:

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

For reasons that will become apparent further down, I find it useful to instead study $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{x^2+y^2}{x-y}\right).$$

Note that for all $(x,y)\in \mathbb R^2$ with $x\neq y$ it holds that $$\dfrac{x^2+y^2}{x-y}=\dfrac{2x^2-(x^2-y^2)}{x-y}=\dfrac{2x^2}{x-y}-(x+y).$$

Since $\lim \limits_{(x,y)\to (0,0)}(x+y)$ exists, any problems that might occur with the initial limit will occur on $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2}{x-y}\right).$$

This last limit is fairly simple to manipulate by choosing $y$ as a function of $x$. If you stare at the fraction long enough, and hopefully not for too long, you'll see that setting $y=x-kx^2$ with $k\in \mathbb R$ will yield a $k$-dependent limit.

You can now abandon this scratch work and consider the paths $t\mapsto (t,t-kt^2)$ in the initial limit.