Manipulating a functional equation 
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 $f((\sqrt 2-1)^3)-f((\sqrt 2+1)^3)$
 A: 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
A: 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.
