1. Let

$$A(x)=\begin{pmatrix} 2\cos^4 x & 2\sin^4 x & 3+\sin^2 2x\\ 3+2\cos^4 x & 2\sin^4 x & \sin^2 2x\\ 2\cos^4 x & 3+2\sin^4 x & \sin^2 2x \end{pmatrix}$$

so that $f(x)=\det A(x)$.

2. To simplify the determinant, use row operations that do not change the determinant:

• $R_2 \to R_2-R_1$
• $R_3 \to R_3-R_1$

Then

$$R_2=(3,0,-3),\qquad R_3=(0,3,-3)$$

Hence

$$f(x)=\begin{vmatrix} 2\cos^4 x & 2\sin^4 x & 3+\sin^2 2x\\ 3 & 0 & -3\\ 0 & 3 & -3 \end{vmatrix}.$$

3. Expand along the first row:

$$f(x)=2\cos^4 x\begin{vmatrix}0 & -3\\ 3 & -3\end{vmatrix} -2\sin^4 x\begin{vmatrix}3 & -3\\ 0 & -3\end{vmatrix} +(3+\sin^2 2x)\begin{vmatrix}3 & 0\\ 0 & 3\end{vmatrix}.$$

Now compute the minors:

$$\begin{vmatrix}0 & -3\\ 3 & -3\end{vmatrix}=0-(-9)=9,$$ $$\begin{vmatrix}3 & -3\\ 0 & -3\end{vmatrix}=-9,$$ $$\begin{vmatrix}3 & 0\\ 0 & 3\end{vmatrix}=9.$$

Therefore,

$$f(x)=2\cos^4 x(9)-2\sin^4 x(-9)+(3+\sin^2 2x)(9).$$

So,

$$f(x)=18\cos^4 x+18\sin^4 x+27+9\sin^2 2x.$$

4. Use the identity

$$\cos^4 x+\sin^4 x=(\cos^2 x+\sin^2 x)^2-2\sin^2 x\cos^2 x =1-2\sin^2 x\cos^2 x.$$

Also,

$$\sin^2 2x=4\sin^2 x\cos^2 x.$$

Thus,

$$\cos^4 x+\sin^4 x=1-\frac{1}{2}\sin^2 2x.$$

Substitute into $f(x)$:

$$f(x)=18\left(1-\frac{1}{2}\sin^2 2x\right)+27+9\sin^2 2x.$$ $$f(x)=18-9\sin^2 2x+27+9\sin^2 2x=45.$$

So $f(x)$ is a constant.

5. Therefore,

$$f'(x)=0 \quad \text{for all } x,$$

and in particular,

$$f'(0)=0.$$

Hence,

$$\frac{1}{5}f'(0)=0.$$

6. Comparing with the options, the correct option is:

$$\boxed{\text{C: }0}$$