(i) To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\) with respect to \(x\).

The derivative of \(2(3x + 4)^{\frac{3}{2}}\) is \(2 \times \frac{3}{2} \times (3x + 4)^{\frac{1}{2}} \times 3 = 9(3x + 4)^{\frac{1}{2}}\).

The derivative of \(-6x\) is \(-6\), and the derivative of \(-8\) is \(0\).

Thus, \(\frac{d^2y}{dx^2} = 9(3x + 4)^{\frac{1}{2}} - 6\).

(ii) To verify the stationary point at \(x = -1\), substitute \(x = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2(3(-1) + 4)^{\frac{3}{2}} - 6(-1) - 8 = 2(1)^{\frac{3}{2}} + 6 - 8 = 0\).

Since \(\frac{dy}{dx} = 0\), there is a stationary point at \(x = -1\).

To determine the nature, evaluate \(\frac{d^2y}{dx^2}\) at \(x = -1\):

\(\frac{d^2y}{dx^2} = 9(3(-1) + 4)^{\frac{1}{2}} - 6 = 9(1)^{\frac{1}{2}} - 6 = 3\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.