$$
\begin{aligned}
& f(x)=x e^{x(1-x)}, x \in R \\
& f^{\prime}(x)=e^{x(1-x)} \cdot\left[1+x-2 x^2\right]
\end{aligned}
$$

$$
\begin{aligned}
& =-e^{x(1-x} \cdot\left[2 x^2-x-1\right] \\
& =-2 e^{x(1-x} \cdot\left[\left(x+\frac{1}{2}\right)(x-1]\right) \\
& f^{\prime}(x)=-2 e^{x(1-x)} A \\
& \text { where } A=\left(x+\frac{1}{2}\right)(x-1)
\end{aligned}
$$
Now, exponential function is always $+$ ve and $f^{\prime}(x)$ will be opposite to the sign of $A$
which is -ve in $\left[-\frac{1}{2}, 1\right]$
Hence, $f^{\prime}(x)$ is $+$ ve in $\left[-\frac{1}{2}, 1\right]$
$\therefore f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$