How find this sum of all distinct values of $f(2014)$

Question

For all functions $f:\mathbb{R}\backslash\{0\}\to\mathbb{R}$, that satisfy

$$f\left(x+\frac1x\right)f\left(x^3+\frac1{x^3}\right) - f\left(x^2+\frac1{x^2}\right)^2 = \left(x-\frac1x\right)^2,$$

find the sum of all distinct values of $f(2014)$.

(2): and Find the possible value $f(2014)$

My idea: let $$x+\dfrac{1}{x}=t,\Longrightarrow x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=t^3-3t$$ so $$f(t)\cdot f(t^3-3t)-f^2(t^2-2)=t^2-4,|t|\ge 2$$

Then I can't Continue,Thank you

Answer 1

If the question has an answer in the sense the number of distinct values $f(2014)$ is at most countable and the sum can be defined, then the answer is $0$.

This is because if $f(x)$ is a solution of the functional equation, so does $-f(x)$.