I am trying to find the local maximum of $\frac{\sqrt{2a}}{a}-\left(E_{\alpha}-E_{\beta}\right)\exp\left(-\frac{a}{4}\right)$, where $E_{\alpha}$ and $E_{\beta}$ are constants:
$$\displaystyle\frac{d}{da}\left[\frac{\sqrt{2a}}{a}-\left(E_{\alpha}-E_{\beta}\right)\exp\left(-\frac{a}{4}\right)\right]=\frac{\sqrt{\frac{a}{2}}-\sqrt{2a}}{a^{2}}+\frac{E_{\alpha}-E_{\beta}}{4}\exp\left(-\frac{a}{4}\right)$$
But when setting this to $0$ to find critical points, I run into problems. I tried taking the constant term $\frac{E_{\alpha}-E_{\beta}}{4}$ onto the LHS and taking the log of both sides. However, I can't seem to isolate $a$. I would grateful for any help.
Let $f(x)=\sqrt{2/x}-4Ke^{-x/4}$ where $K$ is a constant. Then $$f'(x)=-\frac1{\sqrt 2x^{3/2}}+Ke^{-x/4}=0\implies K\sqrt 2x^{3/2}e^{-x/4}=1$$ so the maximum is one of the solutions of $4K^4x^6e^{-x}=1\implies x=-6W\left(-\frac1{6(2K^2)^{1/3}}\right)$, and substitute $K=(E_\alpha-E_\beta)/4$. This simplifies a bit to give $x=-6W\left(-\frac1{3(E_\alpha-E_\beta)^{2/3}}\right)$.
For a more detailed explanation of how the Lambert $W$ function is derived, let us solve the more general equation $ax^be^{-x}=1$. Write $$a^{1/b}xe^{-x/b}=1\implies a^{1/b}\left(-\frac xb\right)e^{-x/b}=-\frac1b\implies\left(-\frac xb\right)e^{-x/b}=-\frac1{ba^{1/b}}.$$ Taking Lambert $W$ of both sides yields $$-\frac xb=W\left(-\frac1{ba^{1/b}}\right)\implies x=-bW\left(-\frac1{ba^{1/b}}\right).$$ In the context of the question, replace $a:=4K^4$ and $b=6$.