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
$$\mathcal{L}=2(\spadesuit)+(\clubsuit)+\ln3\left(\diamondsuit\right).$$
