The maximum value of $y = x{\left(\log x\right)}^2$ is

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

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)$ min or max

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