I have an issue regarding the correct methodology to solve a functional equation.
The aim is to find all the differentiable functions defined as $f:\,]0,+\infty[ \rightarrow R$ such as : $$ (E) :f'(x)f \left( \frac{1}{x} \right) = \frac{1}{x} $$
Personnal attempt :
Supposing $f$ is a solution, we can assume that $f$ is differentiable therefore continuous. We can then differentiate $(E)$ and we get : $$ (E)' : f''(x)f \left( \frac{1}{x} \right)- \frac{1}{x^2}f'(x)f' \left( \frac{1}{x} \right) = -\frac{1}{x^2} $$
If we pose $x = \frac{1}{x}$ we can obtain (supposing that $f' \ne 0$) :
$$ \frac{1}{x}\frac{f''}{f'} - \frac{1}{x} \frac{f'}{f} = -\frac{1}{x^2} $$
Therefore, we can integrate the equation, then apply an exponential and we end up with a 1-st order differential equation.
Book's solution :
They start by substituting $x = \frac{1}{x}$ then, they get the equation :
$$ f' \left( \frac{1}{x} \right)f(x) = x $$
After that, they suppose :
$$ g(x) = f \left( \frac{1}{x} \right)f(x) $$
By differentiating, we get $g'(x) = 0$ therefore : $g(x) =f \left( \frac{1}{x} \right)f(x)= C$ which means we can write:
$$ f'(x) = \frac{\frac{1}{x}}{f \left( \frac{1}{x} \right)} = \frac{\frac{1}{x} f(x)}{f \left( \frac{1}{x} \right)f(x)} = \frac{1}{xCf(x)} $$
which also leads to a 1-st order differential equation but this time : it's different than the one that I personally obtain.
Question
My method seems to be false but I don't understand why : is there a convenient methodology (steps-by-step) for solving functional equations that leads to avoid these kind of misunderstanding ?
Thank you in advance for your help,
Best regards.