Find the inverse and the asymptotic expansion of a logarithmic equation $x(y)=y \ln(y)$ Concerning the logarithmic equation  $x(y)=y\ln(y)$, is there an explicit expression for its inverse function $y=y(x)$?  Mathematica returns $y=\frac{x}{\text{ProductLog}[x]}$, which should be wrong since its limit for $x\to 0^-$ is $1$ rather than the correct answer $0^+$. 
I am specifically interested in the behavior $x\to 0^-$ and $y\to 0^+$, and looking for an asymptotic expansion. Implied by iteration, I guess that: 
$y=\frac{x}{\ln(-x)}+o[\frac{x}{\ln^2(-x)}]$ 
Is this true? What exactly are next orders?

Edit
One more related question: how do Mathematica obtain the form $\frac{x}{\text{ProductLog}[x]}$ (here $\text{ProductLog}$ is Lambert-W function as some answers already pointed out)? 
What I've tried is to use substitution $y=\exp(t)$ and obtained $y=\exp[W(x)]$. These two forms are the same. While I still would like to want to know how.

$y\log{y}$ is not injective on $(0,1)$, so Mathematica is giving you one branch of the inverse, which starts at $x=-1/e$, $y=1/e$ and then increases monotonically through $y=1$ at $x=0$ and so on. You can access the other branch, which starts at the same point and decreases to $x=0$, $y=0$ by replacing ProductLog[x] with ProductLog[-1,x].
It's quite difficult to approximate this as $x \uparrow 0$: the DLMF offers
$$Wm(x) = -\log{(-1/x)}-\log{\log{(-1/x)}}-\frac{\log{\log{(-1/x)}}}{\log{(-1/x)}} + \dotsb, $$
where $Wm$ is what Mathematica calls ProductLog[-1,#]&. One could then divide $x$ by this and use the Taylor expansion of $1/(1-z)$ to get the next couple of terms in the series you are interested in, but it doesn't seem like a good idea to complicate this expansion further!
