We have a function of $x$, which is given by \begin{equation} f(x)=\frac{x^2}{e^x-1} \end{equation} where $x \geq 0$.
Question: Does there exist a closed-form expression for the optimal $x$ that achieves the maximum of $f(x)$?
My approach
Let the derivative of $f(x)$ with respect to $x$ equal $0$: \begin{equation} \begin{aligned} \frac{2x(e^x-1)-x^2e^x}{(e^x-1)^2}&=0 \\ \Rightarrow 2x(e^x-1)-x^2e^x&=0 \qquad (1)\\ \Rightarrow e^x(2-x)&=2 \end{aligned} \end{equation} It seems that the above equation $(1)$ does not have a closed-form solution.
The only non trivial explicit solution of $$e^x(2-x)=2$$ is given in terms of Lambert function $$x_*=2+W\left(-\frac{2}{e^2}\right)\sim 1.59362$$ $$f(x_*)=-W\left(-\frac{2}{e^2}\right) \left(W\left(-\frac{2}{e^2}\right)+2\right)\sim 0.64761$$