The value of the trilogarithm at $\frac{1}{2}$

Differentiating the left side, we find \begin{align} &\frac{d}{dz}\Bigl(\mathrm{Li}_3(z)+\mathrm{Li}_3(1-z)+\mathrm{Li}_3(1-z^{-1})\Bigr)=\\&= \frac{\mathrm{Li}_2(z)-\mathrm{Li}_2(1-z^{-1})}{z}-\frac{\mathrm{Li}_2(1-z)+\mathrm{Li}_2(1-z^{-1})}{1-z}\tag{1} \end{align} Let us now prove that \begin{align} &\mathrm{Li}_2(z)-\mathrm{Li}_2(1-z^{-1})=-\ln z\ln(1-z)+\frac12\ln^2 z+\zeta(2),\tag{2}\\ &\mathrm{Li}_2(1-z)+\mathrm{Li}_2(1-z^{-1})=-\frac12\ln^2z.\tag{3} \end{align} In both cases it suffices to differentiate both sides with respect to $z$ (recall that $\displaystyle\mathrm{Li}_2'(z)=-\frac{\ln(1-z)}{z}$), check that the corresponding expressions coincide and, finally, check the values at $z=1$ to fix the integration constants (e.g. $\zeta(2)$ in (2) appears as $\mathrm{Li}_2(1)$).

Hence we have shown that \begin{align} &\frac{d}{dz}\Bigl(\mathrm{Li}_3(z)+\mathrm{Li}_3(1-z)+\mathrm{Li}_3(1-z^{-1})\Bigr)=\frac{\zeta(2)-\ln z\ln(1-z)}{z}+\frac{\ln^2 z}{2z(1-z)}.\tag{4} \end{align} It is easy to check that the right side of (4) coincides with the derivative of the right side of the identity quoted in the question. Therefore to prove that identity, it now suffices to check it at one point (and $z=1$ is the easiest choice once again).