= Solution
There are two positive roots for $u>1$: the <derivative> of $xe^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the <exponential function> in $1/x$ gives
$$
xe^{1/x}=x+1+\frac1{2x}+O(x^{-2}),
$$
so <series reversion> yields
$$
\boxed{x_{\rm large}=e^u-1+O(e^{-u}).}
$$
For the smaller root, set $y=1/x$ and take a <logarithm>. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ and
$$
\boxed{x_{\rm small}=\frac1u-\frac{\log u}{u^2}+O\left(\frac{(\log u)^2}{u^3}\right).}
$$
These are the first two nonzero terms of the two <asymptotic expansions>. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the <real branches of Lambert W>.
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