First, find the first derivative \(f'(x)\):

\(f'(x) = \frac{d}{dx}[2x + (x + 1)^{-2}] = 2 - 2(x + 1)^{-3}\).

Next, find the second derivative \(f''(x)\):

\(f''(x) = \frac{d}{dx}[-2(x + 1)^{-3}] = 6(x + 1)^{-4}\).

Evaluate \(f'(x)\) at \(x = 0\):

\(f'(0) = 2 - 2(1)^{-3} = 0\).

This shows that \(x = 0\) is a stationary point.

Evaluate \(f''(x)\) at \(x = 0\):

\(f''(0) = 6(1)^{-4} = 6 > 0\).

Since \(f''(0) > 0\), the function has a minimum at \(x = 0\).

(i) Differentiate \(y = \frac{8}{x} + 2x\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}(8x^{-1}) + \frac{d}{dx}(2x) = -8x^{-2} + 2\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(-8x^{-2}) = 16x^{-3} = \frac{16}{x^3}\).

(ii) To find stationary points, set \(\frac{dy}{dx} = 0\):

\(-\frac{8}{x^2} + 2 = 0 \Rightarrow 2x^2 - 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).

For \(x = 2\), \(y = \frac{8}{2} + 2(2) = 4 + 4 = 8\).

For \(x = -2\), \(y = \frac{8}{-2} + 2(-2) = -4 - 4 = -8\).

Check the nature of each stationary point using \(\frac{d^2y}{dx^2}\):

At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{16}{2^3} = 2 > 0\), hence a minimum.

At \(x = -2\), \(\frac{d^2y}{dx^2} = \frac{16}{(-2)^3} = -2 < 0\), hence a maximum.

Given \(u = 2x(y - x)\) and \(x + 3y = 12\).

First, express \(y\) in terms of \(x\):

\(x + 3y = 12\)

\(3y = 12 - x\)

\(y = \frac{12 - x}{3}\)

Substitute \(y\) in the expression for \(u\):

\(u = 2x\left(\frac{12 - x}{3} - x\right)\)

\(u = 2x\left(\frac{12 - x - 3x}{3}\right)\)

\(u = 2x\left(\frac{12 - 4x}{3}\right)\)

\(u = \frac{2x(12 - 4x)}{3}\)

\(u = \frac{24x - 8x^2}{3}\)

\(u = 8x - \frac{8x^2}{3}\)

To find the stationary value, differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 8 - \frac{16x}{3}\)

Set \(\frac{du}{dx} = 0\):

\(8 - \frac{16x}{3} = 0\)

\(8 = \frac{16x}{3}\)

\(24 = 16x\)

\(x = \frac{3}{2}\)

Substitute \(x = \frac{3}{2}\) back into the equation for \(y\):

\(y = \frac{12 - \frac{3}{2}}{3}\)

\(y = \frac{21}{6}\)

\(y = \frac{7}{2}\)

Substitute \(x = \frac{3}{2}\) and \(y = \frac{7}{2}\) back into the expression for \(u\):

\(u = 2 \times \frac{3}{2} \times \left(\frac{7}{2} - \frac{3}{2}\right)\)

\(u = 3 \times 2\)

\(u = 6\)

(i) To find stationary points, set the derivative \(\frac{dy}{dx} = 3x^2 + 2px\) to zero:

\(3x^2 + 2px = 0\)

\(x(3x + 2p) = 0\)

Thus, \(x = 0\) or \(x = -\frac{2p}{3}\).

For \(x = 0\), \(y = 0\), so one stationary point is \((0, 0)\).

For \(x = -\frac{2p}{3}\), substitute back into the equation:

\(y = \left(-\frac{2p}{3}\right)^3 + p\left(-\frac{2p}{3}\right)^2\)

\(y = -\frac{8p^3}{27} + \frac{4p^3}{9} = \frac{4p^3}{27}\)

So the other stationary point is \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\).

(ii) To determine the nature, find the second derivative:

\(\frac{d^2y}{dx^2} = 6x + 2p\)

At \((0, 0)\), \(\frac{d^2y}{dx^2} = 2p > 0\), so it is a minimum.

At \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\), \(\frac{d^2y}{dx^2} = -2p < 0\), so it is a maximum.

(iii) For the curve \(y = x^3 + px^2 + px\), the derivative is:

\(\frac{dy}{dx} = 3x^2 + 2px + p\)

Using the discriminant \(b^2 - 4ac\) for \(3x^2 + 2px + p = 0\):

\((2p)^2 - 4 \cdot 3 \cdot p < 0\)

\(4p^2 - 12p < 0\)

\(4p(p - 3) < 0\)

Thus, \(0 < p < 3\).

To find \(f'(x)\), we use the quotient rule for differentiation. The function \(f(x) = \frac{15}{2x+3}\) can be rewritten as \(f(x) = 15(2x+3)^{-1}\).

The derivative of \(f(x)\) is:

\(f'(x) = -15(2x+3)^{-2} \cdot 2 = \frac{-30}{(2x+3)^2}\).

Since \((2x+3)^2\) is always positive, \(f'(x) < 0\) for all \(x\) in the domain \(0 \leq x \leq 6\). This means \(f(x)\) is strictly decreasing, and therefore, \(f\) has an inverse.