jmc

We first simplify the expressions for $f(x)$ and $g(x).$ We have $$f(x)=\frac{1}{x+\frac{1}{x}}=\frac{1}{\frac{x^2+1}{x}}=\frac{x}{x^2+1}$$and $$g(x)=\frac{1}{x-\frac{1}{x}}=\frac{1}{\frac{x^2-1}{x}}=\frac{x}{x^2-1}.$$Therefore, $$\begin{array}{rl}(g(x){)}^2-(f(x){)}^2& =\frac{x^2}{(x^2-1{)}^2}-\frac{x^2}{(x^2+1{)}^2}\\ & =\frac{x^2\left((x^2+1{)}^2-(x^2-1{)}^2\right)}{(x^2-1{)}^2(x^2+1{)}^2}\\ & =\frac{x^2(4x^2)}{(x^4-1{)}^2}\\ & =\frac{4x^4}{(x^4-1{)}^2}.\end{array}$$Therefore, we have the equation $$\frac{4x^4}{(x^4-1{)}^2}=\frac{5}{4}.$$Cross-multiplying and expanding, we get $$16x^4=5x^8-10x^4+5,$$so $$0=5x^8-26x^4+5.$$This factors as $$0=(5x^4-1)(x^4-5),$$so either $x^4=\frac{1}{5}$ or $x^4=5.$ It follows that the largest solution to the equation is $x=\sqrt[4]{5},$ so $x^2=\boxed{\sqrt{5}}.$