Asymptotics of the inverse of an analytic equation What is an asymptotics for the solution of the analytic (transcendental) equation
$$e^{-x^2}P\Big(\frac1x \Big)=\frac1y$$
where $P$ is a polynomial, as $y\rightarrow\infty$? Would the Lagrange inversion help? One difficulty of applying the Lagrange inversion theorem is that the $e^{-x^2}$ has an essential singularity at $x=\infty$.
As the first step, set $P(u) = u$.
 A: Here is the solution for the simplest case of $P(u)=u$.
Take logarithm of the original equation
$$x = \sqrt{\ln y-\ln x}.\tag1$$
Obviously $x>1$ for large $y$. So $x<\sqrt{\ln y}$ and $x=O(\sqrt{\ln y})$. Taking logarithm
$$\ln x = O(\ln\ln y) \tag2$$
Substitute Equation (2) into Equation (1)
$$x=\sqrt{\ln y}\Big(1+O\Big(\frac{\ln\ln y}{\ln y}\Big)\Big).$$
This is not good enough, since when substituting this expression into the origin equation we do not get the right hand side explicitly. So we iterate Equation (1) and inserting Equation (2) into it. We have
\begin{align}
x &= \sqrt{\ln y - \frac12\ln\big( \ln y-\ln x\big)} \\
&= \sqrt{\ln y - \frac12\ln\big( \ln y-O(\ln\ln y)\big)} \\
&= \sqrt{\ln y - \frac12\ln\ln y-\frac12\ln\Big(1- O\Big(\frac{\ln\ln y}{\ln y}\Big)\Big)} \\
&= \sqrt{\ln y - \frac12\ln\ln y+O\Big(\frac{\ln\ln y}{\ln y}\Big)} \\
& = \sqrt{\ln y - \frac12\ln\ln y}\,\Big(1+O\Big(\frac{\ln\ln y}{(\ln y)^2}\Big)\Big) 
\end{align}
We can continue this iteration indefinitely.

For the general polynomial case we can use sandwiching inequalities
$$e^{-x^2}\frac{1-\delta}x<\frac1y = e^{-x^2}\Big(\frac1x+\frac a{x^2}\Big)<e^{-x^2}\frac{1+\delta}x$$
for large enough $y$ and thus $x$. Then proceed either with the iteration procedure above or the Lambert W function as described in @AntonioVargas' comment above.
