$xf(f(f(x)))=1$ continuous Assume $f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ is a continuous function such that $xf(f(f(x)))=1$ for all $x>0$. 
I found that $f(x)=1/x$ is a solution. Could we find another such function? And why?
 A: No, $f(x)=1/x$ is the only solution.
The proof consists of many simple steps.
part 1: $f$ is strictly decreasing:
First, we show that $f$ is surjective.
Let $x>0$. Then
$$
 f(f(f(1/x)))(1/x)=1
 \quad\Rightarrow\quad
 f(f(f(1/x)))=x.
$$
Thus $f$ is surjective.
Next, we show that $f$ is injective.
Let $x,y>0$ with $f(x)=f(y)$.
Then
$$
 x f(f(f(y))) = 
 x f(f(f(x))) = 1
 = y f(f(f(y)))
$$
which implies $x=y$.
Thus $f$ is injective and therefore bijective.
Since $f$ is bijective and continuous, it has to be either strictly increasing or strictly decreasing.
Assume that $f$ is strictly increasing.
Since the composition of strictly increasing functions is strictly increasing,
it follows that $f\circ f\circ f$ is strictly increasing.
Since $x\mapsto x$ is strictly increasing and the multiplication of strictly increasing nonnegative functions
is strictly increasing, it follows that
$ x \mapsto x f(f(f(x)))=1$ is strictly increasing.
This is a contradiction.
Thus, $f$ has to be strictly decreasing.
part 2: an indirect proof:
Assume that for some $x>0$ we have $f(x)>1/x$.
In the following calculation, each step is either a simple application of $f$
or a multiplication/division by $x$.
Since $f$ is strictly decreasing, it means we have to switch the sign each time we apply $f$.
We obtain
$$
\begin{align}
 && f(x) &>1/x
 \\
 \Rightarrow && f(f(x)) &< f(1/x)
 \\
 \Rightarrow && f(f(f(x)))  &> f(f(1/x))
 \\
 \Rightarrow && 1 = xf(f(f(x)))  &> xf(f(1/x))
 \\
 \Rightarrow && 1/x  &> f(f(1/x))
 \\
 \Rightarrow && f(1/x)  &< f(f(f(1/x)))
 \\
 \Rightarrow && (1/x)f(1/x)  &< (1/x)f(f(f(1/x))) = 1
 \\
 \Rightarrow && f(1/x) &<x
\end{align}
$$
With this information we can start again.
$$
\begin{align}
 && f(x) &>1/x
 \\
 \Rightarrow && f(f(x)) &< f(1/x)
 \\
 \Rightarrow && f(f(x))  &< x
 \\
 \Rightarrow && f(f(f(x)))  &> f(x) > 1/x
 \\
 \Rightarrow && 1=xf(f(f(x))) & > xf(x) > x/x = 1
\end{align}
$$
which is a contradiction.
Now assume $f(x)<1/x$ for some $x>0$.
Then we can obtain a contradiction with the same argument as above,
but exchanging the signs $<$ and $>$.
Thus the only possible option is that $f(x)=1/x$ for all $x$.
