Let f(x)= vmatrix 3x^2 and x and - x 6 and -1 and 0 q and q^2 and...

Given the function $f(x)$ as a determinant where only the first row contains the variable $x$:

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

To differentiate the determinant with respect to $x$, we differentiate the elements of the first row while keeping the other rows unchanged.

Differentiating once with respect to $x$:

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

Differentiating again with respect to $x$:

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

Substituting $x=0$ into the second derivative:

$f''(0)=\begin{vmatrix} 6 \text{ and } -1 \text{ and } 0 \\ 6 \text{ and } -1 \text{ and } 0 \\ q \text{ and } q^2 \text{ and } q^3 \end{vmatrix}$

Since the first and second rows of the determinant are identical, the value of the determinant is zero.

$f''(0) = 0$

Answer: $0$