If $f(x)=x e^{x(1-x)}$, then $f(x)$ is

$ f(x)=x e^{x(1-x)} $ $ f'(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2x) $ $ =e^{x(1-x)}[1+x-2x^2] $ $ e^{x(1-x)}\neq0 $ $ \Rightarrow 1+x-2x^2=0 $ $ \Rightarrow (1-x)(2x+1)=0 $ $ \Rightarrow x=1,\ -\frac{1}{2} $ For $ x\in\left[-\frac{1}{2},1\right],\ f'(x)>0 $ $ \therefore f(x)$ is increasing on $\left[-\frac{1}{2},1\right]$