First of all, I believe the power on the RHS should be $ 2^n$ instead of $ 2n$. We notice that both LHS and RHS can be written as fraction of two polynomials. Hence, it suffices to show the equality holds for infinitely many imaginary numbers. If we let $ x=i\cot(\theta)$, we can see that $ f^{(n)}(x)=i\cot(2^n\theta)$. It is not hard to verify the equality then.

A less creative solution would be: let $ a = f^n(x)$ and $ b = (\frac {x + 1}{x - 1})^{2n}$. Then $ f^{n + 1}(x) = \frac {a^2 + 1}{2a}$. After some calculations, we discover the wanted result is equivalent to $ a^2 = (\frac{b + 1}{b - 1})^2.$ Then we can prove that $ a = \frac {b + 1}{b - 1}$. But this can be proved by induction, since $ x = \frac {y + 1}{y - 1} \Longrightarrow f(x) = \frac {y^2 + 1}{y^2 - 1}$ (which is easily obtained with direct calculation)