Since $\log 2 \approx 0.693$ and $\sin 0=0$, we have $f(0)=\log 2 > 0$.

Near $x=0$: $f(x)=\log 2-\sin x$ (since $\log 2>\sin x$ locally). So $f$ is smooth near $x=0$.

$g(x) = f(f(x)) = |\log 2 - \sin(f(x))|$. At $x=0$: $f(0)=\log 2$, so $\sin(f(0))=\sin(\log 2)$. Since $\log 2 < 1 < \pi$, $\sin(\log 2)>0$.

$g(0) = \log 2 - \sin(\log 2) > 0$, so near $x=0$, $g(x)=\log 2 - \sin(f(x))$.

$g'(0)=-\cos(f(0))\cdot f'(0)=-\cos(\log 2)\cdot(-\cos 0)=\cos(\log 2)$