Find the interval in which $y={{x}^{2}}{{e}^{-x}}$ is an increasing function.

Hint: We need to find the derivative form of the given function $y={{x}^{2}}{{e}^{-x}}$ to find ${{f}^{'}}\left( x \right)>0$.
The interval in which the inequation satisfies ${{f}^{'}}\left( x \right)>0$ will be the interval in which $y={{x}^{2}}{{e}^{-x}}$ is an increasing function. We find the derivation with the help of by product rule. We find the solution of the inequation to solve the problem.

Complete step by step answer:
The given function is $y={{x}^{2}}{{e}^{-x}}$. The function will be an increasing function if we find that ${{f}^{'}}\left( x \right)>0$. This means the second derivative of the curve will have to be positive.
We find the interval in which ${{f}^{'}}\left( x \right)>0$ happens.
We take the derivative of the function $y={{x}^{2}}{{e}^{-x}}$ by differentiating both sides with respect to x. So, \[{{f}^{'}}\left( x \right)=\dfrac{d\left( y \right)}{dx}=\dfrac{d\left( {{x}^{2}}{{e}^{-x}} \right)}{dx}\].
We use the formula of by products we solve the equation
\[{{f}^{'}}\left( x \right)={{e}^{-x}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+{{x}^{2}}\dfrac{d\left( {{e}^{-x}} \right)}{dx}=2x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}\]
We need to find out the interval of x for which ${{f}^{'}}\left( x \right)>0$.
This means $2x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}>0$. We know that for any values of x, \[{{e}^{-x}}\] will be positive.
$\begin{align}
  & 2x{{e}^{-x}}-{{e}^{-x}}{{x}^{2}}>0 \\
 & \Rightarrow {{e}^{-x}}\left( 2x-{{x}^{2}} \right)>0 \\
 & \Rightarrow \left( 2x-{{x}^{2}} \right)>0 \\
\end{align}$
Now we solve quadratic inequation of x to find the interval.
$\begin{align}
  & \left( 2x-{{x}^{2}} \right)>0 \\
 & \Rightarrow \left( {{x}^{2}}-2x \right)<0 \\
 & \Rightarrow x\left( x-2 \right)<0 \\
\end{align}$
From the factorisation we break the interval region as $x\in \left( 0,2 \right)$.

So, the interval in which $y={{x}^{2}}{{e}^{-x}}$ is an increasing function is $\left( 0,2 \right)$.

Note: We need to remember the function becomes increasing when the slope of the curve is greater than 0. This means for every increasing value of x; we find an increasing value of $f\left( x \right)$. The value of \[{{e}^{-x}}\] is always positive as we are taking the power value of a positive and that can never be changed into negative unless we change the base value. That’s why we took out \[{{e}^{-x}}\] part of the equation in the first step.