The function $$f(x) = xe^{x(1-x)}$$, $$x \in \mathbb{R}$$, is

We have $$f(x) = xe^{x(1-x)}$$. To find where this function is increasing or decreasing, we compute the derivative.

Since $$e^{x(1-x)} > 0$$ always, the sign of $$f'(x)$$ is determined by $$-(2x+1)(x-1)$$.

The critical points are $$x = -\frac{1}{2}$$ and $$x = 1$$.

For $$x < -\frac{1}{2}$$: $$(2x+1) < 0$$ and $$(x-1) < 0$$, so $$(2x+1)(x-1) > 0$$, hence $$f'(x) < 0$$ (decreasing).

For $$-\frac{1}{2} < x < 1$$: $$(2x+1) > 0$$ and $$(x-1) < 0$$, so $$(2x+1)(x-1) < 0$$, hence $$f'(x) > 0$$ (increasing).

For $$x > 1$$: $$(2x+1) > 0$$ and $$(x-1) > 0$$, so $$(2x+1)(x-1) > 0$$, hence $$f'(x) < 0$$ (decreasing).

So the function is increasing on $$\left(-\frac{1}{2}, 1\right)$$.

Checking the options: Option 1 says increasing in $$\left(-\frac{1}{2}, 1\right)$$, which matches perfectly.

Hence, the correct answer is Option 1.