Let $ a=\frac{x^{2}}{x-1}$, $ b=\sqrt{x-1}$ and $ c=\frac{\sqrt{x-1}}{x^{2}}$. Then $ abc=1$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. So $ a+b+\frac{1}{ab}=\frac{1}{a}+\frac{1}{b}+ab\Rightarrow (a-1)(b-1)(ab-1)=0$. So now we just have to solve $ a=1$, $ b=1$ and $ c=1$. Solving each equation and noting $ x>1$ is a real, we easily get $ x=2$ as the only solution.