Suggestion for a functional equation $f'\left(\frac{a}{x}\right)=\frac{x}{f(x)}$ where $f:(0,\infty)\to(0,\infty)$ is differentiable I want to solve the following functional equation:

Find all differentiable functions $f : (0,\infty) \rightarrow (0,\infty)$ for which there is a positive real number $a$ such that $$f'\left(\frac{a}{x}\right)=\frac{x}{f(x)}$$ for all $x > 0$.

I have noted that the function $f$ is increasing but i cannot go any further
Any suggestions?
 A: We have
$$f(x)=\frac{x}{f'(a/x)}.$$
Taking derivative wrt $x$, we get
$$f'(x)=\frac{1}{f'(a/x)}-\frac{x}{\big(f'(a/x)\big)^2}\Biggl(-\frac{a}{x^2}f''\left(\frac{a}{x}\right)\Biggr).\tag{1}$$
Plugging in $a/x$ for $x$ in the original functional equation, we get
$$f'(x)=\frac{a/x}{f(a/x)}.\tag{2}$$
From (1) and (2), we have
$$\frac{x}{f(x)}=\frac{1}{f'(x)}+\frac{xf''(x)}{\big(f'(x)\big)^2}$$
for all $x>0$.  This shows that
$$x\big(f'(x)\big)^2=f(x)f'(x)+xf(x)f''(x).$$
So, we have
$$\frac{d}{dx}\left(\frac{f(x)}{f'(x)}\right)=\frac{\big(f'(x)\big)^2-f(x)f''(x)}{\big(f'(x)\big)^2}=\frac{f(x)}{xf'(x)}.\tag{3}$$
Let $g(x)=\frac{f(x)}{f'(x)}$.  Then, (3) is equivalent to
$$\frac{d}{dx}\frac{g(x)}{x}=0.$$
That is, $g(x)=cx$ for some constant $c$.  So $g(x)=cx$, so $\frac{f(x)}{f'(x)}=cx$, so
$$f'(x)=\frac{1}{cx}f(x).$$
This implies
$$\frac{d}{dx}\frac{f(x)}{x^{1/c}}=0.$$
That is, $f(x)=bx^{1/c}$ for some constant $b$.
Now, $f'\left(x\right)=\frac{b}{c}x^{\frac{1-c}{c}}$, so
$$\frac{x}{bx^{1/c}}=\frac{x}{f(x)}=f'\left(\frac{a}{x}\right)=\frac{b}{c}\left(\frac{a}{x}\right)^{\frac{1-c}{c}}=\frac{x}{\frac{a^{\frac{c-1}{c}}c}{b}x^{1/c}}.$$
Therefore,
$$b=\frac{a^{\frac{c-1}{c}}c}{b}.$$
This yields
$$b=a^{\frac{c-1}{2c}}c^{\frac12}.$$
So all solutions are
$$f(x)=a^{\frac{c-1}{2c}}c^{\frac12}x^{\frac1c}$$
for some $c>0$.  If we write $\beta=\frac1c$, we get a simpler form:
$$f(x)=\sqrt{\frac{a^{1-\beta}}{\beta}}x^{\beta}.$$
A: It seems that $f \approx x^\beta $ might work.
