Functional System of Equation

$f(a)+f(b)=f(ab)$

$\displaystyle\rightarrow\,a+b-\frac{1}{a}-\frac{1}{b}=ab-\frac{1}{ab}$

$\rightarrow\,(ab-1)(a-1)(b-1)=0,---(i)$ $ab\neq0$ since domain of $f(x)$ excludes ${0}$ .

$a=f(b)$

$\displaystyle\rightarrow\,a=b-\frac{1}{b}---(ii)$

CASE I: $a=1$ Using this in $(ii)$ , we get $b^2-b-1=0$ .

So the solutions are $\displaystyle\left(1,\frac{1\pm\sqrt5}{2}\right)$ .

CASE II: $b=1$ Using this in $(ii)$ we get $a=0$ which is out of domain, so no solutions.

CASE III: $ab=1$ Using this in $(ii)$ we get $b^2=2\rightarrow\,b=\pm\sqrt2$ . So the solutions are $\displaystyle\left(\frac{1}{\sqrt2},\sqrt2\right),\left(\frac{-1}{\sqrt2},-\sqrt2\right)$ .

So required sum is $\displaystyle\left(1+\frac{1+\sqrt5}{2}\right)+\left(1+\frac{-1+\sqrt5}{2}\right)+\left(\frac{1}{\sqrt2}+\sqrt2\right)+\left(\frac{1}{\sqrt2}+\sqrt2\right)=2+\sqrt5+3\sqrt2=\sqrt4+\sqrt5+\sqrt{18}$ . So we have answer as $4+5+18=\boxed{27}$ .