= Solution
For real $x$, the function $F(x)=x^2e^{-x}$ has <derivative> $F'(x)=x(2-x)e^{-x}$. It decreases from infinity to zero on the negative half-line, increases from zero to $4e^{-2}$ on $(0,2)$, and then decreases to zero. Hence sufficiently small positive $\epsilon$ gives exactly three real roots: two near zero and one tending to positive infinity.
Set $\delta=\sqrt\epsilon$ and write either small root as $x=s\delta+A\delta^2+O(\delta^3)$, with $s=\pm1$. Expanding $x^2e^{-x}$ gives $\delta^2+s(2A-1)\delta^3+O(\delta^4)$. Matching its value to $\delta^2$ fixes $A=1/2$. Thus the two <asymptotic expansions> are
$$
\boxed{x_-=-\sqrt\epsilon+\frac\epsilon2+O(\epsilon^{3/2}),\qquad x_+=\sqrt\epsilon+\frac\epsilon2+O(\epsilon^{3/2}).}
$$
For the large root, put $L=\log(1/\epsilon)$. Taking logarithms gives $x-2\log x=L$, so $x/L\to1$. Writing $x=L+r$ then gives $r=2\log L+2\log(1+r/L)$. Iteration yields
$$
\boxed{x_{\rm large}=L+2\log L+O\!\left(\frac{\log L}{L}\right).}
$$
Its two <asymptotic expansion> terms are of different logarithmic orders; a regular power series in $\epsilon$ cannot describe it.
As an exact check, the <real branches of Lambert W> give $x_-=-2W_0(\sqrt\epsilon/2)$, $x_+=-2W_0(-\sqrt\epsilon/2)$ and $x_{\rm large}=-2W_{-1}(-\sqrt\epsilon/2)$. If complex roots are intended instead, there are infinitely many, given by $-2W_j(\pm\sqrt\epsilon/2)$. For each fixed nonprincipal branch their two logarithmic terms follow from $W_j(z)=L_j-\log L_j+O(\log|L_j|/|L_j|)$, where $L_j=\log z+2\pi ij$ and logarithms are continued on that branch. The three real roots are the ones selected above.
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