Equations (145)-(148) of this paper claim that under certain conditions the function $y(\epsilon)$, given by solving $f(y)=\epsilon$ where $f(y)=a\left(\frac{b}{\sqrt{y}}\right)^y$, has the following asymptotic behaviour:
$$ y(\epsilon) = f^{-1}(\epsilon) = O\left(1+\frac{\log(a/\epsilon)}{\log\log(a/\epsilon)}\right), $$
which follows from the solution in terms of the $W$ function, $y(\epsilon)= \frac{1}{2} W\left(\frac{2 \log \left(\frac{a}{\epsilon }\right)}{b^2}\right)$.
I made a plot of $\frac{\frac{1}{2} W\left(\frac{2 \log \left(\frac{a}{\epsilon }\right)}{b^2}\right)}{1+\frac{\log(a/\epsilon)}{\log\log(a/\epsilon)}}$ for large values of $\xi=1/\epsilon$ but it doesn't seem to approach any constant value, in fact it seems to diverge. Indeed, Mathematica says that the small $\epsilon$ limit of this ratio diverges.
How do I make sense of this statement? Also, how do I calculate the constant prefactors in this expression? (I would like to go beyond Big O.)
$$a \left(\frac{b}{\sqrt{y}}\right)^y=\epsilon\quad \implies \quad y(\epsilon)=b^2 e^{W(t)}\quad \quad\quad \text{where} \quad t=\frac 2{b^2} \log \left(\frac{a}{\epsilon }\right)$$ which is diverging when $\epsilon \to 0$ which means $t \to \infty$.
If you use the expansion $$W(t)=\log (t)-\log (\log (t))+\frac{\log (\log (t))}{\log (t)}+\cdots$$
$$y(\epsilon)\sim b^2\, t\, \Big(\log (t)\Big) ^{\frac{1}{\log (t)}-1}$$ which is a quite good approximation.
Edit
If you look at this paper, $$\log \left(\frac{y(\epsilon )}{b^2}\right)>\log (t)-\log (\log (t))+\frac{\log (\log (t))}{2 \log (t)}$$ $$\log\left(\frac{y(\epsilon )}{b^2}\right)<\log (t)-\log (\log (t))+\frac{e }{e-1 }\frac{\log (\log (t))}{ \log (t)}$$