jmc

Putting $\frac{x+y}{x-y}+\frac{x-y}{x+y}$ over a common denominator, we get $$\frac{2x^2+2y^2}{x^2-y^2}=1.$$Then $2x^2+2y^2=x^2-y^2,$ so $x^2=-3y^2.$

Then $$\begin{array}{rl}\frac{x^4+y^4}{x^4-y^4}+\frac{x^4-y^4}{x^4+y^4}& =\frac{9y^4+y^4}{9y^4-y^4}+\frac{9y^4-y^4}{9y^4+y^4}\\ & =\frac{10}{8}+\frac{8}{10}\\ & =\frac{5}{4}+\frac{4}{5}\\ & =\boxed{\frac{41}{20}}.\end{array}$$