There is no real differentiable function $f$ with $f(f(x))=-x$ As described in the title, I’m trying to prove that there’s no real differentiable function $f$ with that property
I think I should differential it first in order to use the restriction in the problem, but i don’t know how to deal with the results $f’(x)f’(f(x))$ in order to get a contradiction
Thanks in advance for anyone’s help

The fact that $$f \circ f = -\mathrm{Id}, $$ implies that $f$ is bijective (this is a more general fact: if $g \circ h = s$ and if $s$ is bijective, then $h$ is injective and $g$ is surjective; the proof is simple and uses only the definitions of injectivity and surjectivity).
As $f$ is also continuous, this implies that $f$ is strictly monotonous (How can I prove that a continuous injective function is increasing/decreasing?).
But then, $f$ is either increasing or decreasing, and so $f \circ f$ will be increasing, and $-\mathrm{Id}$ is not, so you get a contradiction.
Note that I did not use the assumption of differentiability, only continuity.
