$\mathrm{The}\mathrm{interval}\mathrm{on}\mathrm{which}\mathrm{the}\mathrm{function}f\left(x\right)=x^2{e}^{-x}\mathrm{is}\mathrm{increasing},\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct option is (0, 2).

To find the interval on which the function $f\left(x\right)=x^2{e}^{-x}$ is increasing, we need to find where its first derivative is positive, i.e., $f^{\text{'}}\left(x\right)>0$.

First, let's find the derivative of $f\left(x\right)$ with respect to $x$ using the product rule:
$$f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[x^2\right]·e^{-x}+x^2·\frac{d}{dx}\left[e^{-x}\right]$$

Calculating the derivatives of the individual components:
$$f^{\text{'}}\left(x\right)=2x{e}^{-x}+x^2\left(-e^{-x}\right)$$

Factoring out the common terms $x{e}^{-x}$:
$$f^{\text{'}}\left(x\right)=x{e}^{-x}(2-x)$$

For the function to be increasing, we require:
$$f^{\text{'}}\left(x\right)>0$$
$$x{e}^{-x}(2-x)>0$$

Since the exponential term $e^{-x}$ is always positive for all real values of $x$, the sign of the derivative depends entirely on the product $x(2-x)$. Thus, we must have:
$$x(2-x)>0$$

Multiplying both sides by -1 changes the direction of the inequality:
$$x(x-2)<0$$

The product is negative when $x$ lies strictly between the roots of the expression, which are 0 and 2. Therefore, the inequality holds when:
$$0<x<2$$

Expressing this in interval notation, the function is increasing on the interval (0, 2).