1. **State the problem:** Differentiate implicitly the function $$y=\frac{\ln((2x+4)^9(4x-2))}{\ln\sqrt[5]{3x-1}}$$ to find $$\frac{dy}{dx}$$. 2. **Rewrite the function for clarity:** $$y=\frac{\ln((2x+4)^9(4x-2))}{\ln((3x-1)^{1/5})}$$ 3. **Simplify logarithms using log properties:** $$\ln((2x+4)^9(4x-2))=\ln((2x+4)^9)+\ln(4x-2)=9\ln(2x+4)+\ln(4x-2)$$ $$\ln((3x-1)^{1/5})=\frac{1}{5}\ln(3x-1)$$ So, $$y=\frac{9\ln(2x+4)+\ln(4x-2)}{\frac{1}{5}\ln(3x-1)}=\frac{9\ln(2x+4)+\ln(4x-2)}{\frac{1}{5}\ln(3x-1)}$$ 4. **Rewrite denominator to multiply:** $$y=\frac{9\ln(2x+4)+\ln(4x-2)}{\frac{1}{5}\ln(3x-1)}=\frac{9\ln(2x+4)+\ln(4x-2)}{1} \times \frac{5}{\ln(3x-1)}=\frac{5(9\ln(2x+4)+\ln(4x-2))}{\ln(3x-1)}$$ 5. **Use quotient rule for differentiation:** If $$y=\frac{u}{v}$$ then $$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$ Here, $$u=5(9\ln(2x+4)+\ln(4x-2))$$ $$v=\ln(3x-1)$$ 6. **Find $$\frac{du}{dx}$$:** $$\frac{d}{dx}[9\ln(2x+4)]=9 \times \frac{1}{2x+4} \times 2=\frac{18}{2x+4}$$ $$\frac{d}{dx}[\ln(4x-2)]=\frac{1}{4x-2} \times 4=\frac{4}{4x-2}$$ So, $$\frac{du}{dx}=5\left(\frac{18}{2x+4}+\frac{4}{4x-2}\right)$$ 7. **Find $$\frac{dv}{dx}$$:** $$\frac{d}{dx}[\ln(3x-1)]=\frac{1}{3x-1} \times 3=\frac{3}{3x-1}$$ 8. **Apply quotient rule:** $$\frac{dy}{dx}=\frac{\ln(3x-1) \times 5\left(\frac{18}{2x+4}+\frac{4}{4x-2}\right) - 5(9\ln(2x+4)+\ln(4x-2)) \times \frac{3}{3x-1}}{(\ln(3x-1))^2}$$ 9. **Final answer:** $$\boxed{\frac{dy}{dx}=\frac{5\ln(3x-1)\left(\frac{18}{2x+4}+\frac{4}{4x-2}\right) - 15\frac{9\ln(2x+4)+\ln(4x-2)}{3x-1}}{(\ln(3x-1))^2}}$$ This expression gives the derivative $$\frac{dy}{dx}$$ implicitly for the given function.