Unwrap It Like A Box

Setting $x=y=2$ gives

$f(2 \times 2)=f(2)+f(2) \implies f(4)=2 f(2)$

Setting $x=4,y=2$ gives:

$f(4 \times 2)=f(4)+f(2) \implies f(8)=f(4)+f(2)$

Combing thus with the above result gives:

$f(8)=2f(2)+f(2)=3f(2)$

$\dfrac{f(8)}{f(2)}=\dfrac{3f(2)}{f(2)}=\boxed{\boxed{3}}$

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For a general solution , I think you can do the following:

Differentiate with respect to $x$ and $y$ to yield the following two equations:

$\begin{aligned} y f'(xy) &=f'(x) \implies f'(xy)=\dfrac{f'(x)}{y}\\ x f'(xy)&= f'(y) \implies f'(xy)=\dfrac{f'(y)}{x} \end{aligned}$

Combining the two results gives:

$\dfrac{f'(y)}{x}=\dfrac{f'(x)}{y} \implies x f'(x)=y f'(y)=k \quad \quad \text{For some constant k}$

The result follows because $x,y$ can take any real values so hence $x f'(x)$ must be a constant.

$x f'(x)=k \implies f'(x)=\dfrac{k}{x} \implies f(x)=k \ln{x} + c$

Putting this result back into the original equation gives:

$k \ln{xy}+c=k \ln{x}+c+k \ln{y}+c \implies 2c=c \implies c=0$

This means the general solution is $\boxed{\boxed{f(x)=k \ln{|x|}}}$ which can also be expressed in the form $f(x)=\log_{a}{|x|}$ .

$\begin{array}{lrl} & f(xy) & = f(x) + f(y) \\ \text{Putting }y = x & \implies f(x\cdot x) & = f(x) + f(x) \\ & f(x^2) & = 2f(x) \\ \text{Putting }y = x^2 & \implies f(x\cdot x^2) & = f(x) + f(x^2) \\ & f(x^3) & = 3f(x) \\ \text{Putting }x = 2 & \implies f(2^3) & = 3f(2) \\ & f(8) & = 3f(2) \\ & \implies \dfrac {f(8)}{f(2)} & = \boxed{3} \end{array}$

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Now, we note that:

$\begin{aligned} f(x\cdot 1) & = f(x) + f(1) \\ \implies f(1) & = 0 \\ f(x) & = f(x) \\ f(x^2) & = 2f(x) \\ f(x^3) & = 3f(x) \\ ... & \quad \ ... \\ \implies f(x^n) & = nf(x) \\ \color{#3D99F6}{\implies f(x)} & \color{#3D99F6}{= \log_a x \quad \small \implies f(1) = \log_a 1 = 0, \ f(x^n) = \log_a x^n = n \log_a x = nf(x)} \\ f(10) & = 123456789 \\ \implies \log_a 10 & = 123456789 \\ \log_{10} a & = \frac 1{123456789} \\ \implies f(x) & = \log_a x = \frac {\log_{10}x}{\log_{10}a} \\ \color{#3D99F6}{\implies f(x)} & \color{#3D99F6}{= 123456789 \log_{10} x} \end{aligned}$

$f(8)\\ =f(4\times2)\\ =f(4)+f(2)\\ =f(2\times 2)+f(2)\\ =f(2)+f(2)+f(2)\\ =3f(2)$

Therefore, $\dfrac{f(8)}{f(2)} = \dfrac{3f(2)}{f(2)} = \boxed{3}$

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Bonus question:

Notice that $f(xy)=f(x)+f(y)$ is the same as the logarithmic property $\log_a xy = \log_a x+\log_a y$

Therefore, we can say that

$f(x)=k\log_a x$

However, note that this is only defined for positive real values of $x$ , that is, $x>0$

To allow $f(x)$ to be defined for negative numbers as well, we can add in the absolute value function:

$f(x)=k\log_a |x|$

where $k,a$ are constants that satisfy $a > 0,\;a \neq 1$ and $k \log_a 10 = 123456789\implies a^{123456789} = 10^k$ . There are many possible ordered pairs of $(k,a)$ that satisfy this, and this function is defined for all real values $x$ except $x=0$ , as required from the question

(Note: A better proof is provided in Sam's solution above)

f(8)=f(4x2)=f(4)+f(2) ............... (1)

f(4)=f(2x2)=f(2)+f(2)=2f(4) ................ (2)

from (1) and (2) f(8)=3f(2) then $\frac{f(8)}{f(2)}$ = $\frac{3*f(2)}{f(2)}$ = 3