Finding function $f(x)$ which satisfy given functional equation

Just as you replaced $x$ by $\frac{1}{x}$, you can replace $x$ by $1-x$. Then you can replace $x$ by $\frac{1}{x}$ in this new equation.

Repeating this process, I believe you will obtain 6 independent equations in the variables $f(z)$ where $z$ takes the values $$x,\frac{1}{x},\frac{1}{1-x},\frac{x}{x-1},\frac{x-1}{x},1-x.$$

These can be solved for $f(x)$.

Carrying out this process gives $$24f(x)=-3x+\frac{3}{x}+\frac{1}{1-x}+\frac{7x}{x-1}+\frac{x-1}{x}-5(1-x)$$$$f(x)=\frac{2x^3+x^2+5x-2}{24x(x-1)}$$

I do not understand the phrase "A detailed canonical answer is required to address all the concerns" since I am unaware of any concerns about the answer given some time ago. However, here are the actual equations and how to solve them.

Let $V = \pmatrix{f(x)\\f(\frac{1}{x})\\f(\frac{1}{1-x})\\f(\frac{x}{x-1})\\f(\frac{x-1}{x})\\f(1-x)},$ $W = \pmatrix{x\\\frac{1}{x}\\1-x\\\frac{x-1}{x}\\\frac{x}{x-1}\\\frac{1}{1-x}\\}$

Then $$ \pmatrix{1&2&0&3&0&0\\2&1&3&0&0&0\\0&0&2&0&3&1\\0&0&0&2&1&3\\3&0&0&1&2&0\\0&3&1&0&0&2}V=W$$ Multiplying on the left by $ \pmatrix{-3&3&-5&1&7&1}$ gives $$24f(x)=\pmatrix{-3&3&-5&1&7&1}W$$ Thus giving the answer noted above.

Subtract twice from what you get from the given equation to get:

$$-f(x) + f\left(\frac{x}{x-1}\right) - 2f\left(\frac{1}{1-x} \right) = \frac{x^2 - 2}{3x}$$

Then in this equation plug $x= \frac {x}{x-1}$ to get:

$$-f\left(\frac {x}{x-1}\right) + f\left(x \right) - 2f\left(1-x\right) = \frac{-x^2+4x-2}{3x(x-1)}$$

Adding the two equations you will get:

$$- 2f\left(\frac{1}{1-x} \right) - 2f(1-x) = \frac{x^3 - 2x^2 + 2x}{3x(x-1)}$$

Substituting $x = 1-x$ you should be able to get:

$$f(x) + f\left(\frac{1}{x} \right) = \frac{-x^3 + x^2 - x +1}{6x(1-x)}$$

Substitute in the main equation to get:

$$f\left(\frac{1}{x} \right) + 3f\left(\frac{x}{x-1}\right) = x + \frac{x^3 - x^2 + x -1}{6x(1-x)}$$

Substitute $x = \frac 1x$ to get:

$$f\left(x\right) + 3f\left(\frac{1}{1-x}\right) = \frac{-x^3 + x^2 + 5x -5}{6x(x-1)}$$

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From now on I will stop writing the RHS, as the caclualtions get messy. We substitute $x = \frac{1}{1-x}$ and subtract three times the new equations from the one above the get:

$$f(x) - 9f\left(\frac{x-1}{x}\right) = \cdots$$

Substitute $x = \frac{x-1}{x}$ and add nine times the new equation to the one above to get:

$$f(x)- 81f\left(\frac{1}{1-x}\right) = \cdots$$

Now combining the last equation with the one just above the line I draw you can get an expression for $f(x)$.