Proof that the inverse of $\frac{x}{\ln x}$ is asymptotically equal to $x\ln x$ The inverse of $\frac{x}{\ln x}$ on $x\gt e$ is exactly $-x\operatorname{W}_{-1}\left(-\frac{1}{x}\right)$ where
$$\operatorname{W}_{-1}(x)=\ln\frac{-x}{-\ln\frac{-x}{-\ln\frac{-x}{\ddots}}}.$$
The nested logarithms converge for $-\frac{1}{e}\le x\lt 0$.
Now $-x\operatorname{W}_{-1}\left(-\frac{1}{x}\right)$ should be asymptotically equal to $x\ln x$ as $x\to\infty$.
Maybe we can get something from
$$y=\frac{x}{\ln x}\implies\ln y=\ln x-\ln\ln x,$$
perhaps $\ln\ln x$ could be asymptotically equal to something simpler but I don't see how.
 A: The inverse has
$$x=\frac y{\ln(y)}$$
$$y=x\ln(y)$$
which has $y\ge e$ for $x\ge e$, so
$$y=x\ln(y)\ge x\ln(e)=x$$
$$y=x\ln(y)\ge x\ln(x)$$
On the other hand, we can see that we have $y\le x^2$ when
$$x=\frac y{\ln(y)}\le\frac{x^2}{\ln(x^2)}=\frac{x^2}{2\ln(x)}$$
$$2\ln(x)\le x$$
is true, which occurs when $x\ge e$, so we have
$$y=x\ln(y)\le x\ln(x^2)=2x\ln(x)$$
$$y=x\ln(y)\le x\ln(2x\ln(x))=x\ln(x)+x\ln(2\ln(x))$$
which are tight enough bounds to show that
$$y\sim x\ln(x)$$
as $x\to\infty$.
A: It should be interesting to look at this recent paper where Ioannis Chatzigeorgiou gives nice bounds
$$-1-\sqrt{2u}-u < W_{-1}\left(-e^{-(u+1)}\right) <-1-\sqrt{2u}-\frac 23u$$
Making $u=-1+\log(x)$ leads to
$$-\log (x)-\sqrt{2} \sqrt{\log (x)-1} < W_{-1}\left(-\frac{1}{x}\right)<-\frac{2 \log (x)}{3}-\sqrt{2} \sqrt{\log (x)-1}-\frac{1}{3}$$ and then the conclusion.
A: Let $y=x/\log x$ so that $x=f(y) $ is the intended inverse. We have to show that $x/(y\log y) \to 1$ as $y\to\infty$.
Let's observe that $$\frac{x} {y} =\log x, \log y =\log x - \log\log x$$ and hence $$\frac{x} {y\log y} =\frac{\log x} {\log x-\log\log x} =\dfrac{1} {1-\dfrac{\log t}{t}}$$ where $t=\log x$. As $y\to\infty $ both $x, t$ also tend to $\infty $ and hence the above fraction tends to $1$ and thus $$x\sim y\log y$$ as desired. 
