If $f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3+2{\cos }^{4}x& 2{\sin }^{4}x& {\sin }^{2}2x\\ 2{\cos }^{4}x& 3+2{\sin }^{4}x& {\sin }^{2}2x\end{array}\right|$ then $\frac{1}{5}{f}^{\text{'}}\left(0\right)$ is equal to ________.

Given: $f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3+2{\cos }^{4}x& 2{\sin }^{4}x& {\sin }^{2}2x\\ 2{\cos }^{4}x& 3+2{\sin }^{4}x& {\sin }^{2}2x\end{array}\right|$

Applying $R_2\to R_2-R_1, R_3\to R_3-R_1$

$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3& 0& -3\\ 0& 3& -3\end{array}\right|$

$\Rightarrow f\left(x\right)=9\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 1& 0& -1\\ 0& 1& -1\end{array}\right|$

$\Rightarrow f\left(x\right)=9\left(2{\cos }^{4}x+2{\sin }^{4}x+3+{\sin }^{2}2x\right)$

Now, differentiating above function we get,

$\Rightarrow f^{\text{'}}\left(x\right)=9\left[8{\cos }^{3}x\left(-\sin x\right)+8{\sin }^{3}x\left(\cos x\right)+2\sin 2x\left(\cos 2x\right)\left(2\right)\right]$

$\Rightarrow f^{\text{'}}\left(0\right)=9\left[0+0+0\right]$

$\Rightarrow f^{\text{'}}\left(0\right)=0$