It is given that, 

${\left(f^{\text{'}}\left(x\right)\right)}^4=16{\left(f\left(x\right)\right)}^2,\forall x\in \left(-1,1\right)$

$\Rightarrow {\left(f^{\text{'}}\left(x\right)\right)}^2=\pm 4f\left(x\right)$

Case $\mathrm{I}$ 

If $f\left(x\right)\ge 0\forall x\in \left(-1,1\right)$

Then, $f^{\text{'}}\left(x\right)=\pm 2\sqrt{f\left(x\right)}$

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{\sqrt{f\left(x\right)}}=\pm 2$

$\Rightarrow 2\sqrt{f\left(x\right)}=\pm 2x, \left[\because f\left(0\right)=0\right]$

$\Rightarrow f\left(x\right)=x^2\forall x\in \left(-1,1\right)$

Case $\mathrm{II}$ 

If $f\left(x\right)<0\forall x\in \left(-1,1\right)$

Then, $f^{\text{'}}\left(x\right)=\pm 2\sqrt{-f\left(x\right)}$

$\therefore f\left(x\right)=-x^2\forall x\in \left(-1,1\right)$

$\therefore f_1\left(x\right)=\left[\begin{array}{l}{x}^2,0\le x<1\\ x^2,-1<x<0\end{array}\right.$

$f_2\left(x\right)=\left[\begin{array}{ll}{x}^2& ,0\le x<1\\ -x^2& -1<x<0\end{array}\right.$

$f_3\left(x\right)=\left[\begin{array}{cc}-x^2& ,0\le x<1\\ x^2& ,-1<x<0\end{array}\right.$

$f_4\left(x\right)=\left[\begin{array}{ll}-x^2& ,0\le x<1\\ -x^2& ,-1<x<0\end{array}\right.$

$f_5\left(x\right)=0$

$f_6\left(x\right)=\left[\begin{array}{ll}0& ,0\le x<1\\ x^2& ,-1<x<0\end{array},\dots \right.$ and so on.

Hence, there are more than $4$ functions possible.