Explanation
$$\textbf{Step -1: Find values of x at which the given function changes its nature.}$$
$$\text{The given function is }y(x)=x^2e^{-x}$$
$$\dfrac{dy}{dx}=x^2\dfrac{d}{dx}e^{-x}+e^{-x}\dfrac{d}{dx}x^2$$
$$=-x^2e^{-x}+2xe^{-x}$$
$$\text{Put }\dfrac{dy}{dx}=0.$$
$$\Rightarrow -x^2e^{-x}+2xe^{-x}=0$$
$$\Rightarrow xe^{-x}(-x+2)=0$$
$$\because e^{-x}>0$$
$$\therefore\text{Either }x=0\text{ or }2-x=0$$
$$\Rightarrow x=0\text{ or }2$$
$$\textbf{Step -2: Find the interval in which the given function is increasing.}$$
$$\text{The points }x=0\text{ and }x=2\text{ divide the real line into three disjoint intervals,}$$
$$\text{i.e., }(-\infty,0),(0,2)\text{ and }(2,\infty)$$
$$\because\text{In the interval }(0,2),\dfrac{dy}{dx}>0.$$
$$\therefore \text{In }(0,2),\text{ the given function is increasing.}$$
$$\textbf{Hence, The correct option is D.}$$
$$\left| dy \right| >\left| dx \right| $$
$$\left| \dfrac { dy }{ dx } \right| >1$$
$${ x }^{ 3 }=12y$$
diff wrto y
$$3{ x }^{ 2 }\dfrac { dx }{ dy } =12$$
$$\dfrac { dx }{ dy } =\dfrac { 12 }{ { 3x }^{ 2 } } =\dfrac { 4 }{ { x }^{ 2 } } $$
$$\left| \dfrac { y }{ { x }^{ 2 } } \right| >1$$
$$\left| { x }^{ 2 } \right| >4$$
$$x>\pm 2$$
$$\left( 2,-2 \right) $$
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