Function such that $f^{(n)}(x) = \frac{x}{f(x)^n}$ 
Let $n$ be a fixed positive integer. Find all function $f:(0, \infty) \to \mathbb{R}$ that can be differentiated $n$ times such that $f^{(n)}(x) = \frac{x}{f(x)^n}$ if $f^{(n)}(x)$ is the $n$-th derivative of $f$.

I tried to differentiate the given identity and I wrote $$f^{(n+1)}(x)=\frac{f(x)^n-nxf(x)^{n-1}f'(x)}{f(x)^{2n}}=\frac{f(x)-nxf'(x)}{f(x)^{n+1}}=\frac{1}{f(x)^n}-\frac{nxf'(x)}{f(x)^{n+1}}$$
I tried to connect this with $f^{(n)}(x)$ but the relations didn't get to anything. Also, if $n=2$, I am not able to find any example of a function $f$ that satisfies the equation.
It appears from the comments that the solutions are very complicated. Does the problem become more easy if we replace $f(x)^n$ with $f^n(x)=(f \circ f \circ...\circ f)(x)$?

Not an answer
(I just didn't have enough room to put all this in a comment.)
If you take $f^n(x)=(f\circ f \cdots \circ f) (x)$ as definition, may something nicer happen? For example, consider $n=2$ and restrict yourself to the case
$$f(x) = Kx^\alpha,$$
$\alpha,K \in \Bbb R$ and $x \in \Bbb R^+$, so that
$$f''(x)=K\alpha(\alpha-1) x^{\alpha-2}$$
and
$$(f\circ f)(x) = K(Kx^\alpha)^\alpha,$$
then your condition becomes
$$K\alpha(\alpha-1)x^{\alpha-2} = \frac{x}{K^{1+\alpha}x^{\alpha^2}}.$$
This forces
$$\alpha -2 = 1-\alpha^2 $$
that is
$$\alpha= -\frac{1}{2}\pm \frac{\sqrt{13}}{2}.$$
Call these two values $\alpha_i$, $i=1,2$. Then for the constant factor we must have
$$K^{2-\alpha_i} = \frac{1}{\alpha_i(\alpha_i-1)}.$$
Since $\alpha_i(\alpha_i-1) >0$ we can choose $K$ to be
$$K_i = \left[\alpha_i(\alpha_i-1)\right]^{-\frac{1}{2-\alpha_i}},$$
with $i=1,2$. Is that any useful? Is this solution extendable to an entire family of functions? Is this process generalizable for $n>2$?
