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

\(\frac{d^2y}{dx^2} = 6(-1)^2 - \frac{4}{(-1)^3} = 6 \times 1 + 4 = 10\).

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

(b) Integrate \(\frac{d^2y}{dx^2}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int (6x^2 - \frac{4}{x^3}) \, dx = 2x^3 + \frac{2}{x^2} + c\).

Substitute \(x = -1\) and \(\frac{dy}{dx} = 0\) to find \(c\):

\(0 = 2(-1)^3 + \frac{2}{(-1)^2} + c \Rightarrow c = 0\).

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (2x^3 + \frac{2}{x^2}) \, dx = \frac{1}{2}x^4 - \frac{2}{x} + k\).

Substitute \(x = -1\), \(y = \frac{9}{2}\) to find \(k\):

\(\frac{9}{2} = \frac{1}{2}(-1)^4 - \frac{2}{-1} + k \Rightarrow k = 2\).

Thus, the equation of the curve is \(y = \frac{1}{2}x^4 - \frac{2}{x} + 2\).

(c) Set \(\frac{dy}{dx} = 0\):

\(2x^3 + \frac{2}{x^2} = 0 \Rightarrow x^5 = -1\).

The only real solution is \(x = -1\), so there are no other stationary points.

(d) At \(x = 1\), \(\frac{dy}{dx} = 4\).

Using the chain rule, \(\frac{dx}{dt} = \frac{dx}{dy} \cdot \frac{dy}{dt} = \frac{1}{4} \times 5 = \frac{5}{4}\).