If $f(x)=x e^{x(1-x)}$, then f(x) is 

increasing on [-1/2, 1]

We have,

$f(x)=x e^{x(1-x)}$

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

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

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

Since $e^{x(1-x)}>0$ for all x. Therefore, signs of f'(x) for different values of x are as shown in figure.

Clearly, f(x) is increasing on [-1/2, 1] and decreasing on $(-\infty,-1 / 2] \cup[1, \infty$.