$\text{ Let }f\left(x\right)=x{e}^{x(1-x)},\text{ then }f\left(x\right)\text{ is }$

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=e^{x(1-x)}+x(1-2x)e^{x(1-x)}\\ =\left(1+x-2x^2\right)e^{x(1-x)}\\ =-(x-1)(2x+1)e^{x(1-x)}\end{array}$

$\begin{array}{l}\text{ Since }{e}^{x(1-x)}>0\text{ for all }x\text{, so }{f}^{\mathrm{\prime }}\left(x\right)>0\text{ if and only if }\\ (x-1)(2x+1)<0\text{ i.e. }-1/2=min(1,-1/2)<x<max(1,-1/2)=1\end{array}$

$\text{ Thus f increases on }[-1/2,1]$