1. **State the problem:** Find the derivative $y'$ of the function $$y = \frac{\ln(3x)}{7x}.$$\n\n2. **Recall the formula:** To differentiate a quotient $\frac{u}{v}$, use the quotient rule: $$y' = \frac{u'v - uv'}{v^2}.$$ Here, $u = \ln(3x)$ and $v = 7x$.\n\n3. **Find derivatives of numerator and denominator:**\n- Derivative of $u = \ln(3x)$: Using chain rule, $$u' = \frac{1}{3x} \times 3 = \frac{1}{x}.$$\n- Derivative of $v = 7x$ is $$v' = 7.$$\n\n4. **Apply the quotient rule:**\n$$y' = \frac{\frac{1}{x} \times 7x - \ln(3x) \times 7}{(7x)^2}.$$\n\n5. **Simplify numerator:**\n$$\frac{1}{x} \times 7x = 7,$$ so numerator becomes $$7 - 7\ln(3x).$$\n\n6. **Simplify denominator:**\n$$(7x)^2 = 49x^2.$$\n\n7. **Final derivative:**\n$$y' = \frac{7 - 7\ln(3x)}{49x^2} = \frac{7(1 - \ln(3x))}{49x^2} = \frac{1 - \ln(3x)}{7x^2}.$$\n\n**Answer:** $$\boxed{y' = \frac{1 - \ln(3x)}{7x^2}}.$$