Functional Equation.

Put $x = 2 , y = 1$
$5g(1) = 5 + g(1) + 5 - 2 \rightarrow 4g(1) = 8 \rightarrow g(1) = 2$
Substitute, $y = \dfrac{1}{x}$
$\therefore g(x)g\left(\dfrac{1}{x}\right) = g(x) + g\left(\dfrac{1}{x}\right)$
$\therefore \left( g(x) - 1 \right)\cdot \left(g\left(\dfrac{1}{x}\right) - 1 \right) = 1$

Let,

$g(x) = 1 + f(x) \rightarrow g\left(\dfrac{1}{x}\right) = 1 + f\left(\dfrac{1}{x}\right)$
$\therefore f(x) \cdot f\left(\dfrac{1}{x}\right) = 1 \rightarrow f(x) = \pm x^{n}$
$\therefore g(x) = 1 \pm x^{n}$
Substituting, $x = 2$ , we get $n = 2$
$\therefore g(x) = 1 + x^{2}$
$g(0) + g(1) + g(2) + g(3) = 4 + 1^{2} + 2^{2} + 3^{2} = 18$ .

tbh I guessed

But I made an educated guess:

Which polynomial equation gives 5 when n=2?

I did some guesswork and I got to the equation: g(n) = n^2 +1

And I solved it from there