A continuous function $f:[1, \infty)\to \mathbb{R}$ such that $2xf(x^2)=(x+1)f(x)$, $\forall x\ge 1$ Let $f:[1, \infty)\to \mathbb{R}$ be a continuous function such that $2xf(x^2)=(x+1)f(x)$, $\forall x\ge 1$. Find $f$.
The official solution considers the auxiliary function $g$ such that  $f(x)=\frac{x-1}{x\ln x}g(x), \forall x >1$ and shows that it has the property that $g(x^2)=g(x)$, $\forall x>1$. Then the solution is really easy and straightforward. But how would one come up with such an auxiliary funcion?
 A: The basic idea of such functional equations is to set $f(x) = h(x) g(x)$ for a special function $h(x)$ that deals with the given terms, then show (pray) that $g(x)$ is a constant. 
There isn't really a hard and fast rule to the general case, and comes  with experience and guessing/knowing what types of functions work. 
If we had $ \frac{ f(x^2 ) } { f(x) } = \frac{1}{2} = \frac{ 1/ \ln x^2 } { 1/ \ln x}$, is where the $ \frac{1 }{ \ln x}$ came in from.
Likewise, if we had $ \frac{ f(x^2) } { f(x) } = x +1 = \frac{ x^2 - 1 }{ x-1}$, is where the $ x-1$ came in from.
Finally, if we had $ \frac{ f(x^2 ) } { f(x) } = \frac{1}{x} = \frac{ 1/x^2} { 1/x} $, is where the $ \frac{1}{x}$ came in from.
Putting this all together, we see that if $ \frac{ f(x^2x) } { f(x) } = \frac{ x+1}{2x}$, the function $ \frac{ x-1}{x\ln x}$ comes into play.
(Of course, it's much easier to do in reverse)
So setting $ f(x) = \frac{ x-1}{x\ln x} g(x) $ means we just need to verify that $g(x)$ is a constant. 

As an example, try finding continuous functions for    


*

*$ f(x^2 ) = (x+2) f(x)$.   

*$ f(2x) = f(x)^2 $,

*$f(2x) = 2 \sqrt{1 - x^2} f(x)$. 

