Step 1: Put $t = \log x$, so $\dfrac{dt}{dx} = \dfrac{1}{x}$. Since $f(x) = \log x$, the composite is $g(x) = f(t) = \log t$.
Step 2: By the chain rule, $g'(x) = \dfrac{1}{t}\cdot\dfrac{dt}{dx} = \dfrac{1}{x\log x}$. Write $h(x) = g'(x) = (x\log x)^{-1}$.
Step 3: Use logarithmic differentiation on $h(x)$. Take logs: $\ln h = -\ln(x\log x) = -(\ln x + \ln(\log x))$.
Step 4: Differentiate both sides with respect to $x$. \[\frac{h'}{h} = -\left(\frac{1}{x} + \frac{1}{x\log x}\right) = -\frac{1}{x}\left(1 + \frac{1}{\log x}\right) = -\frac{\log x+1}{x\log x}\]
Step 5: So $h' = h\cdot\left(-\dfrac{\log x+1}{x\log x}\right) = \dfrac{1}{x\log x}\cdot\left(-\dfrac{\log x+1}{x\log x}\right) = -\dfrac{\log x+1}{(x\log x)^2}$, which is exactly $-(x\log x)^{-2}(\log x+1)$. This confirms option (B).
\[\boxed{g''(x) = -(x\log x)^{-2}(\log x+1)}\]