Manipulating a functional equation

Question

Consider a function $f$ such that $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ then find $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$$

And the options are as follows

A) $2f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$

B) $2f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$

C) $f\left(\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )- f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$

D) $f(3)+f\left (\frac {1}{3}\right )$

Through some algebraic manipulations I have reached until the following equations 1) $f(y)\,[ f(y^2)- 1]\, =f(y^3)$

2) $f(1)=0$ or $f(1)=2$

3) $ (f(y)\,)^2- f(1)= f(y^2)$

Also the question can be simplified as to finding $f((\sqrt 2-1)^3)-f((\sqrt 2+1)^3)$

Answer 1

By the properties of multiplication from this equation: $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ we obviously have: $$f\left (\frac {x}{y}\right )=f\left (\frac {y}{x}\right )$$ which means that for any $x$: $$f\left ( x \right )=f\left (\frac {1}{x}\right )$$

Since the arguments in:

$$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right ) $$

are reciprocal of each other, the quantity in question is 0, which makes C) the correct answer

Answer 2

Presumably, it's given that $f$ is non-constant, else we can have $f = 0$ or $f=2$.

So suppose $f$ is non-constant.

Choosing $x$ such that $f(x) \ne 0$, and letting $y=1$, we get $f(1) = 2$.

Letting $x=1$, we get $f(y) = f\left(\frac{1}{y}\right)$.

Noting that $$ \left(\frac{3-2\sqrt 2}{\sqrt 2 + 1}\right )\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right) = 1 $$ it follows that $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right ) = 0$$

By the same reasoning, choice $(C)$ is correct.