1. Given function

We need to study monotonicity of $f(x)=x e^{x(1-x)} = x e^{x-x^2}.$

2. Differentiate the function

Using the product rule, $f'(x)=e^{x-x^2}+x e^{x-x^2}(1-2x).$

Factor out $e^{x-x^2}$: $f'(x)=e^{x-x^2}\left[1+x(1-2x)\right].$

So, $f'(x)=e^{x-x^2}(1+x-2x^2).$

3. Find the sign of $f'(x)$

Since $e^{x-x^2}>0 \quad \text{for all } x\in \mathbb R,$ the sign of $f'(x)$ depends only on $1+x-2x^2.$

Now factor it: $1+x-2x^2=-(2x^2-x-1)=-(2x+1)(x-1)=(1-x)(2x+1).$

Thus, $f'(x)=e^{x-x^2}(1-x)(2x+1).$

So critical points are at $x=-\frac12,\quad x=1.$

4. Check intervals

We study the sign of $(1-x)(2x+1).$

• For $x< -\frac12$: $1-x>0, \quad 2x+1<0 \implies f'(x)<0.$ So $f$ is decreasing on $(-\infty,-\tfrac12)$.

• For $-\frac12 < x < 1$: $1-x>0, \quad 2x+1>0 \implies f'(x)>0.$ So $f$ is increasing on $(-\tfrac12,1)$.

• For $x>1$: $1-x<0, \quad 2x+1>0 \implies f'(x)<0.$ So $f$ is decreasing on $(1,\infty)$.

5. Evaluate the options

• A: increasing in $\left(-\frac12,1\right)$
  This is true.

• B: decreasing in $\left(\frac12,2\right)$
  On $(\tfrac12,1)$, $f'(x)>0$; on $(1,2)$, $f'(x)<0$. Not decreasing on the whole interval.
  False.

• C: increasing in $\left(-1,-\frac12\right)$
  Here $f'(x)<0$, so it is decreasing.
  False.

• D: decreasing in $\left(-\frac12,\frac12\right)$
  Here $f'(x)>0$, so it is increasing.
  False.

6. Final answer

The only correct option is $\boxed{A}.$

7. Comparison with stored answer

Stored correct answer: $A$
Derived answer: $A$
Hence, they agree.