$\begin{align}  & y={{({{x}^{2}}+1)}^{3}} \\ & \text{Let }u=({{x}^{2}}+1) \\ & y={{u}^{3}},\text{ }\frac{dy}{du}=3{{u}^{2}} \\ & \frac{du}{dx}=2x \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=3{{u}^{2}}\times 2x=6x{{u}^{2}} \\ & \frac{dy}{dx}=6x{{({{x}^{2}}+1)}^{2}} \\\end{align}$