I need to show that $\exists\epsilon>0$ s.t. $\forall\delta>0$, $0<\|(x,y)\|<\delta$ and $|\frac{x^2y^2}{x^4+y^4}-L|\geq\epsilon$ for all $L\in\mathbb{R}$.
Now I try to find such an $\epsilon$, but am unable to do so, and am not sure if I am even using the correct approach to proceed.
To find this $\epsilon$, I try to fix arbitrary $\delta$ and $(x,y)$, and see if I can derive some $\epsilon$ which will satisfy the first statement but cannot figure out how.
This problem, where you are (in effect) trying to show that the negation of the $\epsilon,\delta$ definition of a limit holds, is easiest to solve, by first attacking the problem with intuition.
As $(x,y) \to (0,0)$, in any small neighborhood of radius $\delta$ around $(0,0)$ there will be at least one value $(x,y)$ such that $y^2 = kx^2$, where $k$ can be any positive real number.
Therefore, examining the problem intuitively, you have to show that as $k$ varies, the function $f(x,y)$ will critically vary.
$$\frac{x^2y^2}{x^4 + y^4} = \frac{x^4(k)}{x^4(1 + k^2)} = \frac{k}{1 + k^2}.\tag1 $$
Thus, (for example)
when $k=1$, you have that $\displaystyle \frac{k}{1 + k^2} = \frac{1}{2}$
and when $k=2$, you have that $\displaystyle \frac{k}{1 + k^2} = \frac{2}{5}.$
Now, consider that $\displaystyle \frac{1}{2} \times \left|\frac{1}{2} - \frac{2}{5}\right| = \frac{1}{20}.$
Therefore, suppose that you take $\epsilon = \frac{1}{20}$.
Then, regardless of how small $\delta > 0$ is taken, you will be able to find $2$ points $(x_1, y_1)$ and $(x_2, y_2)$ within the neighborhood of radius $\delta$ around $(0,0)$ such that
$|f(x_1,y_1) - f(x_2, y_2)| = 2\epsilon.$
This is game over, by the triangle inequality, regardless of what candidate value of $L$ is suggested.
That is
$$ |f(x_1,y_1) - L| + |f(x_2,y_2) - L| \geq |f(x_1,y_1) - f(x_2, y_2)| = 2\epsilon.\tag1 $$
Therefore, at least one of the two LHS terms in (1) above must be $\geq \epsilon ~:~$ game over.