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