A function f(x) is given such that it satisfies all the real values of x, and a relation is given as,

$$f\left(x\right)\ +\ 2f\left(\dfrac{1}{1-x}\right)\ =\ x$$

What is the value of $$f\left(2\right)$$?

We are given the equation,

$$f\left(x\right)\ +\ 2f\left(\dfrac{1}{1-x}\right)\ =\ x$$

Substituting the value of x = 2 in the equation, we get,

$$f\left(2\right)\ +\ 2f\left(-1\right)\ =\ 2$$   -----(1)

Substituting x = -1, we get

$$f\left(-1\right)\ +\ 2f\left(\dfrac{1}{2}\right)\ = -1$$   -----(2)

Substituting x = $$\dfrac{1}{2}$$ we get,

$$f\left(\dfrac{1}{2}\right)\ +\ 2f\left(2\right)\ =\ \dfrac{1}{2}$$   -----(3)

By applying  (1) - 2*(2) + 4*(3), we get,

$$f\left(2\right)\ +\ 2f\left(-1\right)\ -\ 2f\left(-1\right)\ -\ 4f\left(\dfrac{1}{2}\right)\ +\ 4f\left(\dfrac{1}{2}\right)\ +\ 8f\left(2\right)\ =\ 2\ -\ 2\times\ \left(-1\right)\ +\ \ 4\times\ \dfrac{1}{2}$$

$$f\left(2\right)+\ 8f\left(2\right)\ =\ 2\ +\ 2\ +\ 2$$

$$9f\left(2\right)\ =\ 6$$

$$f\left(2\right)\ =\ \dfrac{2}{3}$$

The correct answer is option A.