先化簡函數 $g(x)$： $$g(x) = \ln \left(2 + \cos^3 x\right)^{\frac{1}{2}} = \dfrac{1}{2} \ln \left(2 + \cos^3 x\right).$$ 利用連鎖律對 $x$ 微分： $$g'(x) = \dfrac{1}{2} \cdot \dfrac{1}{2 + \cos^3 x} \cdot \dfrac{d}{dx}\left(2 + \cos^3 x\right)$$ $$= \dfrac{1}{2} \cdot \dfrac{3\cos^2 x \cdot (-\sin x)}{2 + \cos^3 x} = \dfrac{-3\cos^2 x \sin x}{2\left(2 + \cos^3 x\right)}.$$ 將 $x = \dfrac{\pi}{3}$ 代入，此時 $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$，$\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$： $$g'\left(\dfrac{\pi}{3}\right) = \dfrac{-3 \left(\dfrac{1}{2}\right)^2 \left(\dfrac{\sqrt{3}}{2}\right)}{2\left(2 + \left(\dfrac{1}{2}\right)^3\right)} = \dfrac{-\dfrac{3\sqrt{3}}{8}}{2\left(2 + \dfrac{1}{8}\right)} = \dfrac{-\dfrac{3\sqrt{3}}{8}}{2 \times \dfrac{17}{8}} = -\dfrac{3\sqrt{3}}{34}.$$ 答案為 $-\dfrac{3\sqrt{3}}{34}$。