(a) Given \(\frac{d^2y}{dx^2} = 6x\), integrate to find \(\frac{dy}{dx}\):

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

Since the curve has a stationary point at \((2, -10)\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(3(2)^2 + c = 0\)

\(12 + c = 0\)

\(c = -12\)

Thus, \(\frac{dy}{dx} = 3x^2 - 12\).

(b) Integrate \(\frac{dy}{dx} = 3x^2 - 12\) to find \(y\):

\(y = \int (3x^2 - 12) \, dx = x^3 - 12x + k\).

Substitute \((2, -10)\) into the equation:

\(-10 = 2^3 - 12 \times 2 + k\)

\(-10 = 8 - 24 + k\)

\(-10 = -16 + k\)

\(k = 6\)

Thus, \(y = x^3 - 12x + 6\).

(c) Set \(\frac{dy}{dx} = 0\) to find other stationary points:

\(3x^2 - 12 = 0\)

\(3x^2 = 12\)

\(x^2 = 4\)

\(x = \pm 2\)

Since \(x = 2\) is already given, consider \(x = -2\):

\(y = (-2)^3 - 12(-2) + 6 = -8 + 24 + 6 = 22\)

So, the point is \((-2, 22)\).

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

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

At \(x = -2\), \(\frac{d^2y}{dx^2} = 6(-2) = -12 < 0\), hence a maximum.

(d) Find the equation of the tangent at the y-intercept \((0, y)\):

\(y = 0^3 - 12 \times 0 + 6 = 6\)

The slope \(\frac{dy}{dx}\) at \(x = 0\) is:

\(\frac{dy}{dx} = 3(0)^2 - 12 = -12\)

Equation of the tangent: \(y - 6 = -12(x - 0)\)

\(y = -12x + 6\)