Equality with dilogarithms During some calculations with definite integrals I happened to get the equality 
\begin{eqnarray}
2\, \textrm{Li}_2(-\frac{1}{2}) - 2 \, \textrm{Li}_2(\frac{1}{4})+ 2\, \textrm{Li}_2(\frac{2}{3})= 
3 \log^2 2 - \log^2 3
\end{eqnarray}
Does this follow from some well known equalities for dilogarithms?
 A: I'm going to use the following 3 identities along with the known value $\text{Li}_{2} \left(\frac{1}{2} \right) = \frac{\pi^{2}}{12} - \frac{1}{2} \log^{2}(2)$:
$$\text{Li}_{2}(1-z) = - \text{Li}_{2} \left(1- \frac{1}{z} \right) - \frac{1}{2} \log^{2} (z) , \quad z \notin (-\infty,0] \tag{1}$$
$$\text{Li}_{2}(z) = - \text{Li}_{2} \left(\frac{1}{z} \right) - \frac{\pi^{2}}{6} - \frac{1}{2} \log^{2}(-z) \tag{2}, \quad z \notin [0,1)$$
$$ \text{Li}_{2}(z) = -\text{Li}_{2}(-z) + \frac{1}{2} \text{Li}_{2}(z^{2}) \tag{3}$$

Using $(1)$, which I think is sometimes referred to as Landen's identity, we have
$$\text{Li}_{2} \left(\frac{2}{3} \right) = \text{Li}_{2} \left(1- \frac{1}{3} \right) = - \text{Li}_{2} (-2) - \frac{1}{2} \log^{2} (3).$$
Then using $(2)$ (i.e., the inversion formula),
$$ \text{Li}_{2} \left(\frac{2}{3} \right) =  \text{Li}_{2} \left(- \frac{1}{2} \right) + \frac{\pi^{2}}{6} + \frac{1}{2} \log^{2}(2) - \frac{1}{2} \log^{2}(3).$$
And using $(3)$ we have
$$ \text{Li}_{2} \left(- \frac{1}{2} \right) = -\text{Li}_{2} \left(\frac{1}{2} \right) + \frac{1}{2} \text{Li}_{2} \left(\frac{1}{4} \right) = -\frac{\pi^{2}}{12} + \frac{1}{2} \log^{2}(2) + \frac{1}{2} \text{Li}_{2} \left(\frac{1}{4} \right).$$
Therefore,
$$ \begin{align} &2 \text{Li}_{2}  \left(- \frac{1}{2} \right) -2 \text{Li}_{2} \left(\frac{1}{4} \right) + 2 \text{Li}_{2} \left(\frac{2}{3} \right) \\ &= 4 \text{Li}_{2} \left(-\frac{1}{2} \right)  - 2 \text{Li}_{2} \left(\frac{1}{4} \right)+ \frac{\pi^{2}}{3} + \log^{2}(2) - \log^{2}(3) \\ &= -\frac{\pi^{2}}{3} + 2 \log^{2}(2) + 2 \text{Li}_{2} \left(\frac{1}{4} \right) - 2 \text{Li}_{2} \left(\frac{1}{4} \right) + \frac{\pi^{2}}{3} + \log^{2}(2) - \log^{2}(3) \\ &= 3 \log^{2}(2) - \log^{2}(3). \end{align}$$
