Notice that (by squaring) $f(x):=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}} \le x$ for all $x \ge 1$, and equality holds only when $x=\frac{\sqrt{5}+1}{2}$. The problem asks to solve $f(f(x))=x$, so $x=\frac{\sqrt{5}+1}{2}$.

Hope this is right

Attachments:
Note_Aug_14__2021-compress0.pdf (164kb)

Symmetric function $f(x):=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}} \le x$ $f(x)$ is strictly increasing. $$f(x)=x$$

Find all real numbers x that satisfies $$\sqrt{x-1}+\sqrt{x^2-1}= \sqrt{2} \\\ x$$Find all real numbers x that satisfies $$\sqrt{x-1}+\sqrt{x^2-1}=x\sqrt{x}$$