To find \(\frac{d^2}{dx^2}(f(x))\) at \(x = 0\) for the given function \(f(x)\), we need to first understand the function and its structure:

The function is given by the determinant:

\(f(x) = \begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ q & q^2 & q^3 \end{vmatrix}\)

To differentiate \(f(x)\) twice, let's first calculate this determinant explicitly.

Using the rule for the determinant of a 3x3 matrix:

\(\text{det} = a(ei-fh) - b(di-fg) + c(dh-eg)\)

The expansion of the determinant is:

\[ \begin{align*} f(x) &= 3x^2 \left( (-1) \cdot q^3 - 0 \cdot q^2 \right) - \cos x \left( 6 \cdot q^3 - 0 \cdot q \right) + (-\sin x) \left( 6 \cdot q^2 - (-1) \cdot q \right) \\ &= 3x^2 (-q^3) - \cos x (6q^3) - \sin x (6q^2 + q) \\ &= -3x^2 q^3 - 6q^3 \cos x - \sin x (6q^2 + q). \end{align*} \]

Now, we shall find the first and second derivatives of \(f(x)\).

The first derivative with respect to \(x\) is:

\[ \begin{align*} f'(x) &= \frac{d}{dx} [-3x^2 q^3] + \frac{d}{dx} [-6q^3 \cos x] + \frac{d}{dx} [-\sin x (6q^2 + q)] \\ &= -6x q^3 + 6q^3 \sin x - \cos x (6q^2 + q). \end{align*} \]

To find the second derivative, \(f''(x)\), we differentiate again:

\[ \begin{align*} f''(x) &= \frac{d}{dx} [-6x q^3] + \frac{d}{dx} [6q^3 \sin x] - \frac{d}{dx} [\cos x (6q^2 + q)] \\ &= -6 q^3 + 6q^3 \cos x + \sin x (6q^2 + q). \end{align*} \]

Now, we evaluate \(f''(x)\) at \(x = 0\):

\[ \begin{align*} f''(0) &= -6 q^3 + 6q^3 \cos(0) + \sin(0) (6q^2 + q) \\ &= -6 q^3 + 6q^3 \cdot 1 + 0 \\ &= -6 q^3 + 6 q^3 \\ &= 0. \end{align*} \]

Therefore, \(\frac{d^2}{dx^2}(f(x))\) at \(x = 0\) is equal to 0.

Correct Answer: \(0\)