For $x>e$, express $x/W(x)$ with respect to $\ln(x)$ and $\ln \ln(x)$, where $W(\cdot)$ is the Lambert-W function.
In Wikipedia, we can find the expression of $W(x)$ with respect to $\ln(x)$ and $\ln \ln(x)$. I am not sure how they got that. I need to do this for $x/W(x)$.
Di you mean "productlog"? The word "polylog" may be an abbreviated form of polylogarithm. In the Wolfram Language $\,\texttt{ProductLog[x]}\,$ is used for $\,W(x).\,$ We have $$\, W(x) = x - x^2 + 3x^3/2! - 16x^4/3! + 125x^5/4! + O(x^5) \,$$ and $$\, x/W(x) = 1 + x - x^2/2 + 2x^3/3 - 9x^4/8 + 32x^5/15 + O(x^6). \,$$ These power series have a limited radius of convergence. The Wikipedia series using $\, L_1:=\ln(x)\,$ and $\, L_2:=\ln\ln(x) \,$ are asymptotic but you can just substitute that series into $\, x/W(x). \,$ Explicity, $$\, x/W(x) = x\left(1/L_1 + L_2/L_1^2 + L_2(L_2-1)/L_1^3 + L_2(2 -5L_2 +L_2^2)/(2L_1^4) + O(1/L_1^5)\right). $$