Let $f:(-5,5)\to R$ be a differentiable function with  $f\left(4\right)=1,f^{\text{'}}\left(4\right)=1,f\left(0\right)=-1\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{f}^{\text{'}}\left(0\right)=1.$ If  $g\left(x\right)={\left(f(2f^2\left(x\right)+2)\right)}^2$, then $g^{\text{'}}\left(0\right)$ equals

$$\begin{gathered} g^{\text{'}}\left(x\right)=2f(2f^2\left(x\right)+2)\times f^{\text{'}}(2f^2\left(x\right)+2)\times 4f\left(x\right).f^{\text{'}}\left(x\right) \\ \therefore g^{\text{'}}\left(0\right)=2f(2f^2\left(0\right)+2)\times f^{\text{'}}(2f^2\left(0\right)+2)\times 4f\left(0\right).f^{\text{'}}\left(0\right) \\ =2f\left(4\right)\cdot f^{\text{'}}\left(4\right)\cdot 4f\left(0\right)f^{\text{'}}\left(0\right) \\ =2\times 4(-1)\left(1\right)=-8 \end{gathered}$$