Answer
$$a=1$$
Work Step by Step
Given $$f(x)=x^{2} e^{-x} $$
Since $f(a) = a^2e^{-a} $, then
\begin{align*}
f^{\prime}(x)&=-x^{2} e^{-x}+2 x e^{-x}\\
&=e^{-x}\left(2 x-x^{2}\right)\\
f'(a) &=(2a - a^2)e^{-a}
\end{align*}
Then the tangent line to $f$ at $x = a$ is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&= f'(a) \\
\frac{y- a^2e^{-a} }{x-a}&= (2a - a^2)e^{-a}\\
y&=\left(2 a-a^{2}\right) e^{-a}(x-a)+a^{2} e^{-a}
\end{align*}
The line passes through $(0,0) $ when
\begin{align*}
\left(2 a-a^{2}\right) e^{-a}(-a)+a^{2} e^{-a}&=0\\
e^{-a}\left(a^{2}-2 a^{2}+a^{3}\right)&=0\\
a^{2} e^{-a}(a-1)&=0
\end{align*}
Then $a=0$ or $ a=1 $ and since $a>0 $, then $a=1$
