There was cubes on the left hand side of the given equation, yet no cube in the equation that needs to be proved. We try to work backwards so that there is cubes in the equation:
[math]
\begin{array}{rcl}
\log \left(\dfrac{a - b}2\right) &=& \dfrac13 (\log a + 2 \log b)\\
3 \log \left(\dfrac{a - b}2\right) &=& \log a + 2 \log b\\
\log \left(\left(\dfrac{a - b}2\right)^3\right) &=& \log a + 2 \log b\\
\log \left((a - b)^3\right) - 3 \log 2 &=& \log a + 2 \log b
\end{array}
[/math]
If we can prove that [imath]\log \left((a - b)^3\right) - 3 \log 2 = \log a + 2 \log b[/imath], then it is done.
We were given [imath]a^3 - b^3 = \cdots[/imath]. The equation involves [imath](a - b)^3[/imath]. This hints us to use the identity [imath]a^3 - b^3 = (a - b)^3 + 3ab(a - b)[/imath]. Can you continue from here?