Positive reals $$x, y$$ satisfy $$x \neq y$$ and $$\frac{x^2 + y^2}{xy} = k$$. If replacing $$x$$ by $$x + y$$ and $$y$$ by $$|x - y|$$ leaves the value of $$k$$ unchanged, then $$k$$ equals

Let $$k = \dfrac{x^2 + y^2}{xy}$$.

After the substitution $$x \to x+y$$, $$y \to |x-y|$$:

$$$k' = \frac{(x+y)^2 + (x-y)^2}{(x+y)\cdot|x-y|} = \frac{2(x^2+y^2)}{|x^2-y^2|}$$$

Setting $$k = k'$$ and cancelling the common factor $$x^2 + y^2$$ (positive since $$x, y > 0$$):

$$$\frac{1}{xy} = \frac{2}{|x^2 - y^2|} \implies |x^2 - y^2| = 2xy.$$$

Why we can assume WLOG $$x > y > 0$$: the expression for $$k$$ is symmetric in $$x, y$$ (swap them and $$k$$ is unchanged), and we are told $$x \neq y$$. So one of them is strictly larger; we relabel that one as $$x$$. This removes the absolute value: $$x^2 - y^2 = 2xy$$.

Divide by $$y^2$$ and let $$t = x/y > 1$$:

$$$t^2 - 2t - 1 = 0 \implies t = 1 + \sqrt{2} \quad (\text{taking the positive root}).$$$

Now $$k = \dfrac{x^2 + y^2}{xy} = t + \dfrac{1}{t} = (1 + \sqrt{2}) + \dfrac{1}{1+\sqrt{2}} = (1+\sqrt{2}) + (\sqrt{2} - 1) = \mathbf{2\sqrt{2}}$$.