A special value polylogarithm identity involving $\text{Li}_3(-1/2),\,\text{Li}_3(-1/3),\,\text{Li}_3(2/3),\,\text{Li}_2(-1/3),\,\text{Li}_2(2/3)$ I've found that
\begin{align}
\mathcal{L} = 2\operatorname{Li}_3\left(-\frac{1}{2}\right)+\operatorname{Li}_3\left(-\frac{1}{3}\right)+2\operatorname{Li}_3\left(\frac{2}{3}\right)+\operatorname{Li}_2\left(-\frac{1}{3}\right) \ln(3)+2\operatorname{Li}_2\left(\frac{2}{3}\right) \ln(3)
\end{align}
equals to
$$
\mathcal{L} = \frac{\pi^2}{3}\ln(2)+\frac{1}{3}\ln^3(2)-\frac{1}{3} \ln^2(3) \ln\left(\frac{27}{8}\right)-\frac{\zeta(3)}{6}.
$$

How could we prove this identity?

A numerical approximation:
$$
\mathcal{L} \approx 1.701652530545172752791574942340971991312113932043\dots
$$
 A: This identity (with $z=\frac13$) implies that
$$\operatorname{Li}_3\left(\frac23\right)+\operatorname{Li}_3\left(-\frac12\right)+\operatorname{Li}_3\left(\frac13\right)=\zeta\left(3\right)-\frac{\pi^2}{6}\ln\frac32+\frac12\ln 3\ln^2\frac32-\frac16\ln^3\frac32. \tag{$\spadesuit$}$$
On the other hand this identity gives
$$\operatorname{Li}_3\left(-\frac13\right)-2\operatorname{Li}_3\left(\frac13\right)=-\frac{\ln^33}{6}+\frac{\pi^2}{6}\ln3-\frac{13}{6}\zeta(3).\tag{$\clubsuit$}$$
Adding $2(\spadesuit)+(\clubsuit)$ gives the trilogarithmic part of your expression. 
The dilogarithmic part can be computed using dilogarithmic identities from the same question. They give
$$2\operatorname{Li}_2\left(\frac13\right)-\operatorname{Li}_2\left(-\frac13\right)=\frac{\pi^2}{6}-\frac{\ln^23}{2},$$
or, equivalently, ($\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(1-z\right)=\frac{\pi^2}{6}-\ln z\ln(1-z)$ with $z=\frac13$)
$$2\operatorname{Li}_2\left(\frac23\right)+\operatorname{Li}_2\left(-\frac13\right)=\frac{\pi^2}{6}+\frac{\ln^23}{2}-2\ln 3\ln\frac32.\tag{$\diamondsuit$}$$
Therefore 
$$\mathcal{L}=2(\spadesuit)+(\clubsuit)+\ln3\left(\diamondsuit\right).$$
