1. Let's start by stating the problem: Find the derivative of the function $$g(t) = \left(1 + \frac{\sin 3t}{3} - 2t\right)^{-1}$$ with respect to $t$. 2. Identify the outer function and inner function: The outer function is $$f(u) = u^{-1}$$ and the inner function is $$u = 1 + \frac{\sin 3t}{3} - 2t$$. 3. Apply the chain rule: $$\frac{dg}{dt} = f'(u) \cdot u'$$. 4. Compute the derivative of the outer function: $$f'(u) = -1 \cdot u^{-2} = -u^{-2}$$. 5. Compute the derivative of the inner function: $$u' = \frac{d}{dt} \left(1 + \frac{\sin 3t}{3} - 2t\right) = 0 + \frac{1}{3} \cdot 3 \cos 3t - 2 = \cos 3t - 2$$. 6. Combine these results: $$\frac{dg}{dt} = -u^{-2} \cdot (\cos 3t - 2) = - \frac{\cos 3t - 2}{\left(1 + \frac{\sin 3t}{3} - 2t\right)^2}$$. 7. Final answer: $$\boxed{g'(t) = - \frac{\cos 3t - 2}{\left(1 + \frac{\sin 3t}{3} - 2t\right)^2}}$$