To find $\frac{dy}{dx}$, we will use the chain rule. Step 1: Rewrite the function using exponent notation. $$y = \frac{12}{\sqrt{3x^3 - 9x^2 + 4x - 8}}$$ $$y = 12(3x^3 - 9x^2 + 4x - 8)^{-\frac{1}{2}}$$ Step 2: Apply the chain rule. Let $u = 3x^3 - 9x^2 + 4x - 8$. Then $y = 12u^{-\frac{1}{2}}$. The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. First, find $\frac{dy}{du}$: $$\frac{dy}{du} = \frac{d}{du}(12u^{-\frac{1}{2}}) = 12 \cdot \left(-\frac{1}{2}\right)u^{-\frac{1}{2} - 1}$$ $$\frac{dy}{du} = -6u^{-\frac{3}{2}}$$ Next, find $\frac{du}{dx}$: $$\frac{du}{dx} = \frac{d}{dx}(3x^3 - 9x^2 + 4x - 8)$$ $$\frac{du}{dx} = 3(3x^{3-1}) - 9(2x^{2-1}) + 4(1x^{1-1}) - 0$$ $$\frac{du}{dx} = 9x^2 - 18x + 4$$ Step 3: Multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$. $$\frac{dy}{dx} = (-6u^{-\frac{3}{2}}) \cdot (9x^2 - 18x + 4)$$ Step 4: Substitute $u = 3x^3 - 9x^2 + 4x - 8$ back into the expression for $\frac{dy}{dx}$. $$\frac{dy}{dx} = -6(3x^3 - 9x^2 + 4x - 8)^{-\frac{3}{2}}(9x^2 - 18x + 4)$$ Step 5: Rewrite the expression with positive exponents and in radical form. $$\frac{dy}{dx} = \frac{-6(9x^2 - 18x + 4)}{(3x^3 - 9x^2 + 4x - 8)^{\frac{3}{2}}}$$ $$\frac{dy}{dx} = \frac{-6(9x^2 - 18x + 4)}{(\sqrt{3x^3 - 9x^2 + 4x - 8})^3}$$ The final answer is $\boxed{\frac{dy}{dx} = \frac{-6(9x^2 - 18x + 4)}{(\sqrt{3x^3 - 9x^2 + 4x - 8})^3}}$.