The equation of the curve is \(y = \frac{k}{x}\). To find the gradient, we differentiate \(y\) with respect to \(x\).

\(\frac{dy}{dx} = -\frac{k}{x^2}\).

We are given that the gradient \(\frac{dy}{dx} = -3\) when \(x = 2\).

Substitute \(x = 2\) into the derivative:

\(-\frac{k}{2^2} = -3\).

Simplify:

\(-\frac{k}{4} = -3\).

Multiply both sides by \(-4\):

\(k = 12\).

To find the gradient of the curve, we need to differentiate \(y = \frac{12}{x^2 - 4x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = 12\) and \(v = x^2 - 4x\).

First, find \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = 2x - 4\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(x^2 - 4x)(0) - 12(2x - 4)}{(x^2 - 4x)^2}\)

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

\(= -12(x^2 - 4x)^{-2} \times (2x - 4)\)

Now, substitute \(x = 3\) into the derivative:

\(\frac{dy}{dx} = -12(3^2 - 4 \times 3)^{-2} \times (2 \times 3 - 4)\)

\(= -12(9 - 12)^{-2} \times (6 - 4)\)

\(= -12(-3)^{-2} \times 2\)

\(= -12 \times \frac{1}{9} \times 2\)

\(= -\frac{24}{9}\)

\(= -\frac{8}{3}\)

(i) To find \(\frac{dy}{dx}\), differentiate \(y = x^2 + \frac{2}{x}\):

\(\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{2}{x}\right)\)

\(\frac{dy}{dx} = 2x - \frac{2}{x^2}\)

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2x - \frac{2}{x^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}\left(-\frac{2}{x^2}\right)\)

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

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

\(2x - \frac{2}{x^2} = 0\)

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

\(2x^3 = 2\)

\(x^3 = 1\)

\(x = 1\)

Substitute \(x = 1\) into \(y = x^2 + \frac{2}{x}\):

\(y = 1^2 + \frac{2}{1} = 3\)

Stationary point is \((1, 3)\).

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

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

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

(i) Differentiate \(y = x^3 + 3x^2 - 9x + k\) with respect to \(x\):

\(\frac{dy}{dx} = 3x^2 + 6x - 9\).

(ii) Stationary points occur where \(\frac{dy}{dx} = 0\):

\(3x^2 + 6x - 9 = 0\).

Factorize the quadratic equation:

\(3(x^2 + 2x - 3) = 0\).

\(3(x + 3)(x - 1) = 0\).

Thus, \(x = -3\) or \(x = 1\).

(iii) For the curve to have a stationary point on the \(x\)-axis, \(y = 0\) at the stationary points:

Substitute \(x = -3\) into \(y = x^3 + 3x^2 - 9x + k\):

\((-3)^3 + 3(-3)^2 - 9(-3) + k = 0\).

\(-27 + 27 + 27 + k = 0\).

\(k = -27\).

Substitute \(x = 1\) into \(y = x^3 + 3x^2 - 9x + k\):

\(1^3 + 3(1)^2 - 9(1) + k = 0\).

\(1 + 3 - 9 + k = 0\).

\(k = 5\).

Thus, the values of \(k\) are \(-27\) and \(5\).

(a) To find \(k\), first find the derivative \(f'(x) = 2x - \frac{k}{x^2}\).

Set \(f'(2) = 0\):

\(2 \times 2 - \frac{k}{2^2} = 0\)

\(4 - \frac{k}{4} = 0\)

\(k = 16\)

(b) To determine the nature of the stationary point, find the second derivative \(f''(x) = 2 + \frac{2k}{x^3}\).

Evaluate \(f''(2)\):

\(f''(2) = 2 + \frac{2 \times 16}{2^3} = 2 + \frac{32}{8} = 6\)

Since \(f''(2) > 0\), the stationary point is a minimum.

(c) When \(x = 2\), \(f(x) = 2^2 + \frac{16}{2} + 2 = 4 + 8 + 2 = 14\).

Since this is the only stationary point and it is a minimum, the range of \(f\) is \(y \geq 14\).