Find all functions $f: \Bbb{R} \to \Bbb{R} $ such that $f(f(x))=-x$ 
Find all functions  $f: \Bbb{R} \to \Bbb{R} $ such that $f(f(x)) = -x$.

I think this equation has no continuous solution, because no polynomial can be the solution and we know that every continuous function is the limit of a sequence of polynomials.
 A: if f(a) = b, then f(b) = -a, f(-a) = -b, and f(-b) = a.
for every ordered pair (a,b) that the function maps, there is a 4 fold symmetry (90 degree rotations), mapping 3 other ordered pairs.
A: Applying $f$ to the equation shows that $f$ is an odd function, in particular $f(0)=0$. 
Assuming differentiable such $f$ existed, we can differentiate at $x=0$ and obtain the contradiction $f'(0)^2=-1$.
Now lets assume that $f$ is merely monotone locally at $0$. If it were monotonously increasing, then we would have $f(x)>0$ for small enough $x>0$ and $f(x)\to 0$ for $x\to 0$. Thus $f(f(x))>0$ for even smaller $x$ by continuitiy, giving a contradiction. If it were monotonously decreasing we would get that $f(x)<0$ for small enough $x>0$ and $f(x)\to 0$ for $x\to 0$. Together with oddness we obtain the contradiction $f(f(x))>0$ again. 
Finally, let's show that continuous such $f$ existed. Then it cannot be monotone, and by the intermediate value theorem $f(x)=f(y)$ for $x\neq y$. Applying $f$ shows $-x=-y$, a contradiction.
FINAL REMARKS: a more succinct version of this answer can be found in one of the MO links in the comments of one of the duplicate links
A: The answer of Doug is useful.
If $c=f(0)$ then $f(c)=f(f(0))=0$ so $-c=f(f(c))=f(0)=c$ showing that $c=0$.
If $x\neq0$ then $f(f(x))=-x\neq x$ so there are no $2$-cykels.
Also $f(f(f(f(x))))=-f(f(x))=x$.
So the function must be bijective and has one $1$-cykel and further has $4$-cykels.
Construct a partition $\mathcal P$ of $\mathbb R$ with $\{0\}\in\mathcal P$ and such that every other element takes the form $\{u,v,-u,-w\}$ with $u,v>0$ and $u\neq v$ 
If $\{u,v,-u,-v\}$ is such an element of the partition then define $f(u)=v,f(v)=-u,f(-u)=-v,f(-v)=u$ or $f(u)=-v,f(-v)=-u,f(-u)=v,f(v)=u$
Then $f$ will be a function with the mentioned property and every function $f$ with that property will take this form.
