If f " ( x ) is a positive function for all x ∈ R , f ' ( 3 ) = 0 and g ( x ) = f tan 2 ( x )…

$f\text{'}\text{'}\left(x\right)$ is a positive function for all $x\in R$,

$\Rightarrow f\text{'}\left(x\right)$ is an increasing function for all $x\in R$

As, $f\text{'}\left(3\right)=0$

$\Rightarrow f\text{'}\left(x\right) >0 ,$ for all $x>3$  $...\left(1\right)$

Now, we have a function $g\left(x\right)=f\left({\tan }^2\left(x\right)-2\tan \left(x\right)+4\right)$

$\Rightarrow g\text{'}\left(x\right)=f\text{'}\left({\tan }^2\left(x\right)-2\tan \left(x\right)+4\right)\times \frac{d\left({\tan }^2\left(x\right)-2\tan \left(x\right)+4\right)}{dx}$

Consider, $g\left(x\right)$ is an increasing function, then $g\text{'}\left(x\right)>0$

$\Rightarrow g\text{'}\left(x\right)=f\text{'}\left({\tan }^2\left(x\right)-2\tan \left(x\right)+4\right)\times \left(2\tan x·{\sec }^{2}x-2{\sec }^{2}x\right)>0$

$\Rightarrow f\text{'}\left({\tan }^2\left(x\right)-2\tan \left(x\right)+4\right)\times \left(2\tan x·{\sec }^{2}x-2{\sec }^{2}x\right)>0$

Now as, ${\tan }^2\left(x\right)-2\tan \left(x\right)+4={\tan }^2\left(x\right)-2\tan \left(x\right)+1+3={\left(\tan x-1\right)}^2+3>3$

Then from condition $\left(1\right)$,

$\Rightarrow 2\tan x·{\sec }^{2}x-2{\sec }^{2}x>0\Rightarrow \tan x>1 \Rightarrow x\in \left(\frac{\mathrm{\pi }}{4}, \frac{\mathrm{\pi }}{2}\right)$