The equation of the curve is \(y = 3x - \frac{4}{x}\).

The derivative is \(\frac{dy}{dx} = 3 + \frac{4}{x^2}\).

The gradient of \(AB\) is \(m_{AB} = 4\).

Set \(\frac{dy}{dx} = 4\) to find \(C\) and \(D\):

\(3 + \frac{4}{x^2} = 4\)

\(\frac{4}{x^2} = 1\)

\(x^2 = 4\)

\(x = \pm 2\)

For \(x = 2\), \(y = 3(2) - \frac{4}{2} = 6 - 2 = 4\), so \(C(2, 4)\).

For \(x = -2\), \(y = 3(-2) - \frac{4}{-2} = -6 + 2 = -4\), so \(D(-2, -4)\).

The midpoint \(M\) of \(CD\) is \((0, 0)\).

The gradient of \(CD\) is \(m_{CD} = 2\).

The perpendicular gradient is \(-\frac{1}{2}\).

The equation of the perpendicular bisector is \(y = -\frac{1}{2}x\).