1. **State the problem:** We need to find the derivative $\frac{dy}{dx}$ of the function $$y = -6 \ln(\ln(\ln(x)))$$. 2. **Recall the chain rule:** For a composite function $y = f(g(x))$, the derivative is $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x).$$ 3. **Apply the chain rule step-by-step:** Let $$u = \ln(x),$$ then $$v = \ln(u) = \ln(\ln(x)),$$ and $$y = -6 \ln(v) = -6 \ln(\ln(\ln(x))).$$ 4. **Differentiate each part:** - Derivative of $y$ with respect to $v$: $$\frac{dy}{dv} = -6 \cdot \frac{1}{v} = -\frac{6}{v}.$$ - Derivative of $v$ with respect to $u$: $$\frac{dv}{du} = \frac{1}{u} = \frac{1}{\ln(x)}.$$ - Derivative of $u$ with respect to $x$: $$\frac{du}{dx} = \frac{1}{x}.$$ 5. **Combine using the chain rule:** $$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} = -\frac{6}{v} \cdot \frac{1}{u} \cdot \frac{1}{x}.$$ 6. **Substitute back $v$ and $u$:** $$\frac{dy}{dx} = -\frac{6}{\ln(\ln(x))} \cdot \frac{1}{\ln(x)} \cdot \frac{1}{x} = -\frac{6}{x \ln(x) \ln(\ln(x))}.$$ **Final answer:** $$\boxed{\frac{dy}{dx} = -\frac{6}{x \ln(x) \ln(\ln(x))}}.$$