$\begin{align}  & y=3{{x}^{3}}-9{{x}^{2}} \\ & \frac{dy}{dx}=9{{x}^{2}}-18x=9x(x-2) \\ & \text{At stationary point }\frac{dy}{dx}=0 \\ & 9x(x-2)=0 \\ & 9x=0\text{ or (}x-2)=0 \\ & x=0\text{ or }x=2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=18x-18 \\ & \text{At }x=0 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=18(0)-18=-18<0\text{  (max}\text{. point)} \\ & \text{At }x=2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=18(2)-18=18>0\text{  (min point)} \\\end{align}$The minimum value of x is 2