$\begin{align}  & y={{\left( x-\frac{1}{x} \right)}^{-3}} \\ & \text{Let }u=\left( x-\frac{1}{x} \right) \\ & y={{u}^{-3}},\text{  }\frac{dy}{du}=-3{{u}^{-4}} \\ & \frac{du}{dx}=1+\frac{1}{{{x}^{2}}} \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=-3{{u}^{-4}}\times \left( 1+\frac{1}{{{x}^{2}}} \right) \\ & \frac{dy}{dx}=-3{{\left( 1-\frac{1}{x} \right)}^{-4}}\left( 1+\frac{1}{{{x}^{2}}} \right)=\frac{-3\left( 1+\frac{1}{{{x}^{2}}} \right)}{{{\left( 1-\frac{1}{x} \right)}^{4}}} \\\end{align}$