Is there a real-valued function $f$ such that $f(f(x)) = -x$? Is there a function $f\colon \mathbb{R} \to\mathbb{R} $ such that $ f(f(x)) = -x$ ?  
 A: There is.
Let $\{A,B\}$ be a partition of the positive reals such that $|A|=|B|=|\mathbb{R}|$, and let $\varphi:A\to B$ be a bijection. Define $f:\mathbb{R}\to\mathbb{R}$ as follows: 
$$f(x)=\begin{cases}
0,&\text{if }x=0\\
\varphi(x),&\text{if }x\in A\\
-\varphi^{-1}(x),&\text{if }x\in B\\
-\varphi(-x),&\text{if }-x\in A\\
\varphi^{-1}(-x),&\text{if }-x\in B\;.
\end{cases}$$
Then 
$$\begin{align*}
f(f(x))&=\begin{cases}
0,&\text{if }x=0\\
f(\varphi(x)),&\text{if }x\in A\\
f(-\varphi^{-1}(x)),&\text{if }x\in B\\
f(-\varphi(-x)),&\text{if }-x\in A\\
f(\varphi^{-1}(-x)),&\text{if }-x\in B\;.
\end{cases}\\\\
&=\begin{cases}
0,&\text{if }x=0\\
-\varphi^{-1}(\varphi(x)),&\text{if }x\in A\\
-\varphi(\varphi^{-1}(x)),&\text{if }x\in B\\
\varphi^{-1}(\varphi(-x)),&\text{if }-x\in A\\
\varphi(\varphi^{-1}(-x)),&\text{if }-x\in B
\end{cases}\\\\
&=-x\;.
\end{align*}$$
The idea is simply that $f$ permutes the sets $A,B,-A$, and $-B$ in the order
$$A\stackrel{f}\longrightarrow B\stackrel{f}\longrightarrow -A\stackrel{f}\longrightarrow -B\stackrel{f}\longrightarrow A$$
while leaving $0$ fixed. (Here $-A= \{-a:a\in A\}$, and similarly for $-B$.)
Added: It is possible, though a bit messy, to define $A,B$ and $\varphi$ explicitly. We may, for example, set $A=(0,1]$ and $B=(1,\to)$ and define $\varphi$ as follows. First, for $n\in\omega$ let $\varphi(2^{-n})=2^{n+1}$, so that $\varphi(1)=2,\varphi(1/2)=4,\varphi(1/4)=8$, and so on. Then let $\varphi$ map the interval $(2^{-(n+1)},2^{-n})$ to the interval $(2^n,2^{n+1})$ in the obvious way, taking $x$ to $1/x$.
