<s>[INST] Difficult Functions Evaluation Problem I have a question about finding the value of a certain function that I cannot wrap my head around. 
The question is:  Given a function $f(x)$ satisfying $$f(x) + 2f\left(\frac{1}{1-x}\right) = x,$$ Then find $f(2).$ 
So far, I have tried plugging 2 into the original equation to yield $f(2) + 2f(-1) = 2.$
Next, I plugged $\displaystyle \frac{1}{2}$ into the original equation to yield $\displaystyle f\left(\frac{1}{2}\right) + 2f(2) = \frac{1}{2}$. 
However, I do not know how to solve this system of equations for $f(2).$ 
Please let me know of any hints you may have. Many thanks.
 [/INST] Given $$f(x)+2f\left(\frac{1}{1-x}\right) = x...................(1)$$
Replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get 
$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right) = \frac{1}{1-x}.......(2)$$
Again replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get 
$$f\left(\frac{1}{1-x}\right)+2f(x) = \frac{x-1}{x}..............(3)$$
Now from $(1)$ and $(2)$
$$f(x)-4f\left(\frac{x-1}{x}\right)=x-\frac{2}{1-x}..........(4)$$
Multiply $(3)$ by $4\;,$ We get
$$4f\left(\frac{1}{1-x}\right)+8f(x) = \frac{4(x-1)}{x}..........(5)$$
Add $(4)$ and $(5)$
$$9f(x) = x-\frac{2}{1-x}+\frac{4(x-1)}{x}$$
So $$9f(2)=2+2+2=6\Rightarrow f(2) = \frac{2}{3}$$
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