Is my proof of $\neg (f(f(x)) = f^{-1}(x)) \land (f(x) = x) $ correct? Theorem: $$ f(f(x)) = f^{-1}(x) \implies f(x) = x, \ \ f : \Bbb R \to \Bbb R $$

Proof:
$f(x)$ is a real, continuous function that satisfies $f(f(x)) = f^{-1}(x)$.
That means that either $f : [x,f(x)] \to [f(x), f^{-1}(x)]$ or $f : [f(x),x] \to [ f^{-1}(x), f(x)]$. This is because the function is strictly monotonic, since it has an inverse.
This interval mapping shows us that $f(x) - x = f^{-1}(x) - f(x)$, which means $2f(x) -x = f^{-1}(x)$, yielding the equality:
$$f(f(x))  = 2f(x) -x$$
From that, we gather some info about the function itself:
$$f(x) = 2x - f^{-1}(x) $$
This leads to the fact of $f(f(x) -2x) = x$, with the consequence of:
$$f(f(f(f(f(x) -2x)))) = f(f(x) -2x) = x$$
This can only be true if $f(x) = x$.

EDIT:
The above proof has been edited a lot. In the first revision, I added a vital component I had forgotten. However, the revised proof was still incorrect, although the conclusion was correct. I have now reworked the proof quite drastically, and I believe it is correct.
 A: As other people have mentioned in the comments, there are several issues with the clarity of your proof. It can be assumed that by $f(x)$ is a function, you mean a continuous real valued function $f:\mathbb{R}\to S\subseteq\mathbb{R}$. I will continue under this assumption.
At the beginning of the second paragraph you are making an unjustified step. Why would $[x,f(x)]$ map to $[f(x),f^{2}(x)]=[f(x),f^{-1}(x)]$? There is a property of $f$ you are trying to use implicitly, but confused slightly:

 that property is that continuous functions with an inverse are monotone

From this you should be able to directly derive some inequalities that show that $f(x)=x.$
Hope this helps.
P.S. It might also help to think of your conditions as saying that
$f^3(x)=x$ AND $f$ is invertible.
A: well, no. $$ f(x) = \frac{-1}{x+1} $$
function composition for Moebius transformations is nothing more complicated then matrix multiplication.
Take
$$
P = 
\left( 
\begin{array}{rr} 
0 & -1 \\ 
1 & 1 \\ 
\end{array}
\right) 
$$
$$
P^3 = 
\left( 
\begin{array}{rr} 
-1 & 0 \\ 
0 & -1 \\ 
\end{array}
\right) 
$$
which gives the identity function $ \frac{-x}{-1} $
