If we know that $$ f'(k/x)f(x) = x\tag{ * } $$ Then what can we say about
$$f(k/x)f'(x) ?$$
Originally I tried substituting $x=k/x$ into (*), to give
$$f'(x)f(k/x) = k/x$$
But is this valid? I'm guessing not. Ultimately I'm trying to find a solution for $f(x)$, and the above doesn't seem to working.
Any thoughts appreciated.
Yes, this variable replacement is correct. Now consider $g(x)=f(x)f\left(\frac kx\right)$. It has derivative $$ g'(x)=f'(x)f\left(\frac kx\right)-f(x)f'\left(\frac kx\right)\frac k{x^2}=\frac kx-\frac kx=0 $$ so that $g(x)=C=const$. Thus the functional differential equation is reduced to a more simple functional equation. $$ f(x)f\left(\frac kx\right)=C=f(\sqrt k)^2. \tag1 $$ This means that the values of $f$ for $x>\sqrt{k}$ are given by the values of $f$ at $x\le\sqrt{k}$. Insert this insight (1) into the original equation, $$ f\left(\frac kx\right)=\frac{C}{f(x)}=\frac{C}{x}\frac{x}{f(x)}=\frac{C}{x}f'\left(\frac kx\right)\implies \frac{f'(x)}{f(x)}=\frac k{Cx}\implies f(x)=\tilde C x^{k/C} $$ so essentially $f(x)=ax^d$. Try another round of confirmations: $$ f'(x)=adx^{d-1}\implies f(x)f'\left(\frac kx\right)=a^2dk^{d-1}x, $$ thus the constants need to be connected by $a^2dk^{d-1}=1$.