Junior Balkan MO

We start with the given equation:

$$\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k.$$

Let's set $a = \frac{x^2+y^2}{x^2-y^2}$ and $b = \frac{x^2-y^2}{x^2+y^2}$. Therefore, we have:

$$a + b = k.$$

Observe that:

$$ab = \left(\frac{x^2+y^2}{x^2-y^2}\right) \left(\frac{x^2-y^2}{x^2+y^2}\right) = \frac{(x^2+y^2)(x^2-y^2)}{(x^2-y^2)(x^2+y^2)} = 1.$$

Hence, $a$ and $b$ are the solutions to the quadratic equation:

$$t^2 - kt + 1 = 0.$$

Next, we need to evaluate:

$$E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{x^8-y^8}{x^8+y^8}.$$

Rewriting the terms involving $x^8$ and $y^8$:

$$\frac{x^8 + y^8}{x^8 - y^8} = \frac{(x^4+y^4)(x^4-y^4) + 2x^4y^4}{(x^4+y^4)(x^4-y^4)} = a',$$

and

$$\frac{x^8 - y^8}{x^8 + y^8} = \frac{(x^4+y^4)(x^4-y^4)}{(x^4+y^4)(x^4-y^4) + 2x^4y^4} = b'.$$

Thus, we need to find an expression for:

$$\frac{x^8 + y^8}{x^8-y^8} - \frac{x^8-y^8}{x^8+y^8} = a' - b'.$$

Recognizing the similarity of these expressions with $a$ and $b$, and given they arise due to the power symmetry, we can assume:

$$a' = a^4 \quad \text{and} \quad b' = b^4.$$

Thus,

$$a' - b' = a^4 - b^4.$$

Given $a + b = k$ and $ab = 1$, by the identity for the fourth powers:

$$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 = (k^2 - 2)^2 - 2,$$

and

$$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (k\sqrt{k^2 - 4})(k^2 - 2).$$

After some algebraic manipulations:

$$a^4 - b^4 = \frac{k^4 - 8k^2 + 16}{4k}.$$

Thus the answer is:

$$\boxed{\frac{k^4 - 8k^2 + 16}{4k(k^2 + 4)}}.$$