Given that $y = \dfrac{3\ln 2x}{x^2}$ for $x>0$.

1. Identify the components for the quotient rule and find their derivatives.

Let $u = 3\ln 2x$ and $v = x^2$.
Then $\dfrac{du}{dx} = 3 \left(\dfrac{1}{2x}\right)(2) = \dfrac{3}{x}$
and $\dfrac{dv}{dx} = 2x$.

2. Apply the quotient rule.

$\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}$
$\dfrac{dy}{dx} = \dfrac{x^2\left(\dfrac{3}{x}\right) - (3\ln 2x)(2x)}{(x^2)^2}$

3. Simplify the expression to obtain the required form.

$\dfrac{dy}{dx} = \dfrac{3x - 6x\ln 2x}{x^4}$
Factorising $3x$ from the numerator:
$\dfrac{dy}{dx} = \dfrac{3x(1 - 2\ln 2x)}{x^4}$
$\dfrac{dy}{dx} = \dfrac{3}{x^3}(1 - 2\ln 2x)$ (Shown)

1. Express integration as the reverse of differentiation using the result from part (a)

Since $\frac{d}{dx}\left(\frac{3\ln 2x}{x^2}\right) = \frac{3}{x^3}(1 - 2\ln 2x)$,
$\int \frac{3}{x^3}(1 - 2\ln 2x)\,dx = \frac{3\ln 2x}{x^2} + C_1$

2. Expand the integral and separate the terms

$\int \frac{3}{x^3}\,dx - \int \frac{6\ln 2x}{x^3}\,dx = \frac{3\ln 2x}{x^2} + C_1$
$3\int x^{-3}\,dx - 6\int \frac{\ln 2x}{x^3}\,dx = \frac{3\ln 2x}{x^2} + C_1$

3. Evaluate the standard integral and rearrange to find the target integral

$3\left(\frac{x^{-2}}{-2}\right) - 6\int \frac{\ln 2x}{x^3}\,dx = \frac{3\ln 2x}{x^2} + C_1$
$-\frac{3}{2x^2} - \frac{3\ln 2x}{x^2} - C_1 = 6\int \frac{\ln 2x}{x^3}\,dx$
$\int \frac{\ln 2x}{x^3}\,dx = -\frac{3}{12x^2} - \frac{3\ln 2x}{6x^2} + C$
$\int \frac{\ln 2x}{x^3}\,dx = -\frac{1}{4x^2} - \frac{\ln 2x}{2x^2} + C$