If f ( x + y ) = f ( x ) f ( y ) and Σ _ x = 1 ^ ∞ f(x) = 2, x, y ∈...

$f ( x ) = {a}^{x}$ satisfies the functional relation, so
$\mathop{Σ}_{x = 1}^{\infty}f\left(x\right) = 2$
$⇒ a + {a}^{2} + {a}^{3} \ldots \ldots \infty \text{terms} = 2$
$⇒ \dfrac{a}{1 - a} = 2$

$⇒ a = \dfrac{2}{3}$
$⇒ f\left(x\right) = {\left(\dfrac{2}{3}\right)}^{x}$
Now, $\dfrac{f ( 4 )}{f ( 2 )} = \dfrac{{\left(\dfrac{2}{3}\right)}^{4}}{{\left(\dfrac{2}{3}\right)}^{2}} = \dfrac{4}{9}$