The minimum value of $\mathrm{y}=\mathrm{x}{(\log \mathrm{x})}^2$ is

$$\begin{gathered} \mathrm{y}=\mathrm{x}{(\log \mathrm{x})}^2 \\ \frac{\mathrm{dy}}{\mathrm{dx}}={(\log \mathrm{x})}^2+\mathrm{x} 2\mathrm{logx}.\frac{1}{\mathrm{x}} \\ =\mathrm{logx}(\log \mathrm{x}+2) \end{gathered}$$

y is min or max

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$$\begin{gathered} \log \mathrm{x}=0, \mathrm{logx}=-2 \\ \mathrm{x}=1 \mathrm{x}={\mathrm{e}}^{-2} \\ \mathrm{x}=1 \Rightarrow \mathrm{y}=1\left(0\right)=0 \\ \mathrm{x}={\mathrm{e}}^{-2}\Rightarrow \mathrm{y}={\mathrm{e}}^{-2}{(-2)}^2 \\ =\frac{4}{{\mathrm{e}}^2} \end{gathered}$$

$\therefore$ min value is 0