Let $f:R\to R$ satisfy  $f\left(x+y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)\forall x,y,\in R$. If $f\left(2\right)=3$ then $7\frac{f^{\text{'}}\left(4\right)}{f^{\text{'}}\left(2\right)}=$  

$f\left(x+y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)$

$puty=2$

$f\left(x+2\right)=2^x.\left(3\right)+16f\left(x\right)$

$f^{\text{'}}\left(x+2\right)={3.2}^x\log 2+16f^{\text{'}}\left(x\right)$

$f^{\text{'}}\left(4\right)=12\log 2+16f^{\text{'}}\left(2\right)\_\_\_\_\_\_\_\_\_\left(1\right)$

put  $x=2$

$f\left(2+y\right)=4f\left(y\right)+4^{y}f\left(2\right)$

$f^1\left(2+y\right)=4f^1\left(y\right)+{3.4}^y\log 4$

$f^1\left(4\right)=4f^1\left(2\right)+48\log 2^2$

$f^1\left(4\right)=4f^1\left(2\right)+96\log 2------\left(2\right)$

From (1) and (2) 

$f^1\left(2\right)=7\log 2$

$f^1\left(4\right)=124\log 2$

$7\frac{f^1\left(4\right)}{f^1\left(2\right)}=\frac{7\left(124\log 2\right)}{7\log 2}=124$