Let us assume the function has a limit $L \in \Bbb{R}$ as $(x, y) \to (0, 0)$. We can use the $\varepsilon$-$\delta$ definition to obtain a contradiction.
How will this work? As per the $\varepsilon$-$\delta$ definition, we know that we can force the function value $f(x, y)$ to be come $\varepsilon$-close to our limit $L$ (and we get total control over what that $\varepsilon > 0$ is, but we'll choose it later), so long as as we take $(x, y)$ sufficiently close (i.e. closer than some $\delta$ that we have no control over) to $(0, 0)$. If we can pick $(x_0, y_0)$ that is within this $\delta$ from $0$ (again, we don't know what $\delta$ is, or have any control over it, but we do get to pick our $(x_0, y_0)$), then we are guaranteed that $|f(x_0, y_0) - L| < \varepsilon$.
Normally, we're interested in ensuring that our $f(x_0, y_0)$ is indeed within $\varepsilon$. Now, it's essentially an assumption. We choose our $(x_0, y_0)$ carefully, and the fact that $|f(x_0, y_0) - L| < \varepsilon$ means that we can figure out roughly where $L$ lies! Specifically, it means that
$$f(x_0, y_0) - \varepsilon < L < f(x_0, y_0) + \varepsilon$$
This is good: we don't know the limit, but we can narrow it down to a small interval. And because we have some control over the value of $f(x_0, y_0)$, and all the control over $\varepsilon$, we get some control over this interval.
To get our contradiction, we just need to narrow $f(x_0, y_0)$ down to being in two small intervals that don't intersect! Which intervals would they be?
You know from your own investigation that, by following different lines through the origin, you get different limits. You might have noticed that following $y = x$, i.e. taking function values of the form $f(x, x)$, that you constantly get $1$. Whereas, if you take $y = 0$, i.e. function values of the form $f(x, 0)$, you constantly get $0$.
If we follow the lines to within distance $\delta$ of $(0, 0)$ (no matter how small $\delta$ ends up being), then we guarantee that $L$ lies within distance $\varepsilon$ of $1$, but also distance $\varepsilon$ of $0$. But, of course, we get to choose $\varepsilon$! What if we made $\varepsilon = 1/2$? We cannot have $L$ be within half distance of $1$ and half distance of $0$. Otherwise, we would have
$$-\frac{1}{2} < L < \frac{1}{2} < L < \frac{3}{2},$$
which would imply $L < L$, a contradiction. Choosing $\varepsilon$ to be half the difference between $1$ and $0$ ensured that our two little intervals, one around $0$, the other around $1$, do not intersect, which gives us our contradiction.
Let's prove it properly:
Proof
Assume $\lim_{(x, y) \to (0, 0)} \frac{2xy}{x^2 + y^2} = L$. Let $\varepsilon = \frac{1}{2}$. By the limit definition, there exists some $\delta > 0$ such that
\begin{align*}
0 < \sqrt{x^2 + y^2} < \delta &\implies |f(x, y) - L| < \varepsilon = \frac{1}{2} \\
&\implies f(x, y) - \frac{1}{2} < L < f(x, y) + \frac{1}{2}.
\end{align*}
Consider $(x_0, y_0) = (\delta/2, 0)$. [This is a "close enough" point along the line $y = 0$.] Note that $f(x_0, y_0) = 0$. Then,
$$0 < \sqrt{x_0^2 + y_0^2} = \sqrt{\frac{\delta^2}{4} + 0} = \frac{\delta}{2} < \delta.$$
By the limit definition, we conclude that
$$f(x_0, y_0) - \frac{1}{2} < L < f(x_0, y_0) + \frac{1}{2},$$
that is
$$-\frac{1}{2} < L < \frac{1}{2}.$$
Similarly, consider now $(x_1, y_1) = (\delta/2, \delta/2)$. [This is a "close enough" point along the line $y = x$.] Note that $f(x_1, y_1) = 1$. Once again, we get
$$0 < \sqrt{x_1^2 + y_1^2} = \sqrt{\frac{\delta^2}{4} + \frac{\delta^2}{4}} = \sqrt{\frac{\delta^2}{2}} = \frac{\delta}{\sqrt{2}} < \delta.$$
Thus, by the limit definition,
$$f(x_1, y_1) - \frac{1}{2} < L < f(x_1, y_1) + \frac{1}{2}$$
i.e.
$$1 - \frac{1}{2} < L < 1 + \frac{1}{2} \iff \frac{1}{2} < L < \frac{3}{2}.$$
But, this is a contradiction. We could previously conclude that $L < \frac{1}{2}$, but we have now just concluded $L > \frac{1}{2}$. Thus, our assumption that the limit existed must be false. $\square$
Best Answer
HINT:
The idea is to start from the $\epsilon-\delta$ definition of continuity. So you write $$f(x+\delta_1,y+\delta_2)-f(x,y)=[f(x+\delta_1,y+\delta_2)-f(x,y+\delta_2)]+[f(x,y+\delta_2)-f(x,y)]$$