Let $g: A \rightarrow A$ be defined by $g(x):=1-1 / x$; the key property is that $g(g(g(x)))=1-\frac{1}{1-\frac{1}{1-\frac{1}{x}}}=x$. The given equation rewrites as $f(x)+f(g(x))=\log |x|$. Substituting $x=g(y)$ and $x=g(g(z))$ gives the further equations $f(g(y))+f(g(g(y)))=\log |g(x)|$ and $f(g(g(z)))+f(z)=\log |g(g(x))|$. Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives $f(x)=\frac{1}{2} \cdot(\log |x|-\log |g(x)|+\log |g(g(x))|)$. For $x=2007$, we have $g(x)=\frac{2006}{2007}$ and $g(g(x))=\frac{-1}{2006}$, so that $f(2007)=\frac{\log |2007|-\log \left|\frac{2006}{2007}\right|+\log \left|\frac{-1}{2006}\right|}{2}=\log (2007 / 2006)$.