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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?

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
$$xbig(f'(x)big)^2=f(x)f'(x)+xf(x)f''(x).$$
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, $f'left(xright)=frac{b}{c}x^{frac{1-c}{c}}$, 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:
$$f(x)=sqrt{frac{a^{1-beta}}{beta}}x^{beta}.$$

It seems that $f approx x^beta $ might work.