What is the value of $x$, given that $f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$ and $f^{-1}\left(x\right)=2$? I tried to approach the problem by using the equation $f\left(2\right)=x$ but I always get stuck in the middle of the process.

We have
$$x=2\Rightarrow f(2)+f(-1)=2$$
$$x=1/2\Rightarrow f(1/2)+f(2)=1/2$$
$$x=-1\Rightarrow f(-1)+f(1/2)=-1$$
This is a system of three equations with three unknowns. If it helps, we can rewrite it as
$$s+y=2$$
$$s+z=1/2$$
$$y+z=-1$$
where $f(2)=s$, $f(-1)=y$, and $f(1/2)=z$. We can easily solve this by noting that
$$2+1/2=(s+y)+(s+z)=2s+(y+z)=2s-1$$
$$7/2=2s$$
$$s=7/4$$
We conclude $f(2)=7/4$ and therefore $x=7/4$.
