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

Now $f\left(x\right)=x{e}^{x\left(1-x\right)}$

Differentiating both side w.r.t $x$ we get,

$f^{\text{'}}\left(x\right)=x{e}^{x\left(1-x\right)}\left(1-2x\right)+e^{x\left(1-x\right)}$

$=-e^{x\left(1-x\right)}\left(2x^2-x-1\right)$

$=-e^{x\left(1-x\right)}\left(2x+1\right)\left(x-1\right)$

Now finding critical point by equating $f^{\text{'}}\left(x\right)=0$

We get, $x=\frac{-1}{2} \& x=1$

$\therefore f\left(x\right)$ is decreasing in $\left(-\infty ,-\frac{1}{2}\right)\cup \left(1,\infty \right)$

And increasing in $\left(-\frac{1}{2},1\right)$