The interval in which

We have,

$y=x^{2} e^{-x}$

$\therefore \frac{d y}{d x}=2 x e^{-x}-x^{2} e^{-x}=x e^{-x}(2-x)$

Now, $\frac{d y}{d x}=0$

$\Rightarrow x=0$ and $x=2$

The points $x=0$ and $x=2$ divide the real line into three disjoint intervals i.e., $(-\infty, 0),(0,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(2, \infty), f^{\prime}(x)<0$ as $e^{-x}$ is always positive.

$\therefore f$ is decreasing on $(-\infty, 0)$ and $(2, \infty)$.

In interval $(0,2), f^{\prime}(x)>0$.

$\therefore f$ is strictly increasing on $(0,2)$.

Hence, f is strictly increasing in interval (0, 2).

The correct answer is D.