Differentiation and Invers become one

$\large F(x)=f\left( e^{g(x)} \right)$

$\large F'(x)=\left(f'\left( e^{g(x)} \right) \right) e^{g(x)} g'(x)$

$\large g(x)=f^{-1}(x) \Rightarrow f\left(g(x)\right)=g\left(f(x)\right)=x$

$\large g\left(f(x)\right)=x$

Differentiating wrt $\large x$ we get,

$\large g'\left(f(x)\right)f'(x)=1$

$\large \Rightarrow g'\left(f(x)\right)=\dfrac{1}{f'(x)} \hspace{4 mm} [1]$

$\large F'(3)=\left(f'\left( e^{g(3)} \right) \right) e^{g(3)} g'(3)$

now its interesting to note that

$\large f(0)=3\Rightarrow g(3)=g\left(f(0)\right)=0$

from $\large [1]$ we have

$\large g'(3)=g'\left(f(0)\right)=\dfrac{1}{f'(0)}$

Substituting these values we get,

$\large F'(3)=f'\left( e^{g\left( f\left( 0 \right) \right)} \right) e^{g\left( f\left( 0 \right) \right)} g'(f(0))=f'(e^0)e^0\dfrac{1}{f'(0)}=\dfrac{f'(1)}{f'(0)}$

$\large f'(x)=3x+5 \Rightarrow\dfrac{f'(1)}{f'(0)}=\dfrac{8}{5}$

Thus $\large a+b=8+5=\boxed{13}$