Solve function equation $$ f ( x y ) = f ( x ) ^ { y ^ \beta } f ( y ) ^ { x ^ \beta } \text , $$ where $ \beta $ is real number, $ f ( x ) $ is continuous single valued function.
Obviously, $ f ( x ) = 1 $ satisfies it, but are there other solutions?
There is a way to answer that question that is due to Enrico Fermi (see E. Fermi, Thermodynamics, Ch.IV). Of course, being him a physicist treating a physical variable like entropy, he made several cavalier assumptions about the $f(x)$ entering into the functional equation as, e.g., that a Taylor expansion exists. If this is the case, being $y$ arbitrary, we will assume $y=1+\epsilon$ and expands on $\epsilon$. You will get the following equality $$ f(x)+\epsilon xf'(x)+O(\epsilon^2)=f(x)f(1)+\epsilon\beta f(1)f(x)\log(f(x))+\epsilon x^\beta f(x)f'(x)+O(\epsilon^2). $$ Then, by equating equal powers of $\epsilon$, it is easy to see that $$ f(1)=1 $$ and $$ xf'(x)=\beta f(x)\log(f(x))+x^\beta f(x)f'(x) $$ that is the differential equation to solve for this problem. But this has just the trivial solution.
First of all, note that there are ambiguities in the statement of the problem that need to be resolved before starting to solve it. The most important one is that the domain and codomain of $ f $ should be specified. Another one is that it should be specified for which elements of the domain like $ x $ and $ y $ we have the functional equation $$ f ( x y ) = f ( x ) ^ { y ^ \beta } f ( y ) ^ { x ^ \beta } \text . \tag 0 \label 0 $$ Note for example that if $ \beta < 0 $, $ 0 ^ \beta $ has no meaning, if $ \beta $ is irrational, $ ( - 1 ) ^ \beta $ is undefined, and exponentiation of complex numbers (to the power of a non-integer) is not a well-defined single-valued function.
Let's assume that $ f : \mathbb R ^ + \to \mathbb R ^ { 0 + } $ and \eqref{0} hold for all $ x , y \in \mathbb R ^ + $. Under this specification, we can show that the only solutions are of the form $ f ( x ) = 0 $ and $ f ( x ) = \exp \big( x ^ \beta A ( \log x ) \big) $ for some additive function $ A : \mathbb R \to \mathbb R $. It's straightforward to check that these are indeed solutions. We prove that they are the only ones.
Define $ g : \mathbb R ^ + \to \mathbb R ^ { 0 + } $ with $ g ( x ) = f ( x ) ^ { x ^ { - \beta } } $. Then, by \eqref{0} we have $$ g ( x y ) = g ( x ) g ( y ) \tag 1 \label 1 $$ for all $ x , y \in \mathbb R ^ + $. Lettin $ x = y = 1 $ in \eqref{1} we can see that $ g ( 1 ) \in \{ 0 , 1 \} $. If $ g ( 1 ) = 0 $, then putting $ y = 1 $ in \eqref{1} shows that $ g $ is constantly zero, and thus $ f $ too is the constant zero function, which is one of the mentioned solutions. So, from now on, assume that $ g ( 1 ) = 1 $. Letting $ y = \frac 1 x $ in \eqref{1} we have $$ g ( x ) g \left( \frac 1 x \right) = 1 \text , $$ which in particular shows that $ g ( x ) > 0 $ for all $ x \in \mathbb R ^ + $. This lets us define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = \log \big( g ( \exp x ) \big) $, and \eqref{1} will show that $ A $ satisfies Cauchy's functional equation. Thus we have $$ f ( x ) = g ( x ) ^ { x ^ \beta } = \Big( \exp \big( A ( \log x ) \big) \Big) ^ { x ^ \beta } = \exp \big( x ^ \beta A ( \log x ) \big) \text , $$ which gives the other mentioned set of solutions.
It's well-known that assuming the axiom of choice, one can show that wild additive functions exists, using Hamel bases of $ \mathbb R $ (considered as a vector space over the field $ \mathbb Q $). On the other hand, assuming further regularities like continuity, local boundedness or even Lebesgue measurability, an additive function $ A : \mathbb R \to \mathbb R $ must be linear; i.e. there must be a constant $ a \in \mathbb R $ such that $ A ( x ) = a x $ for all $ x \in \mathbb R $. See this post for more information. In the case of our problem, assuming regularities on $ f $ forces regularities on $ A $, and thus regular nonzero solutions to \eqref{0} must be of the form $ f ( x ) = \exp \left( a x ^ \beta \log x \right) $, or equivalently $$ f ( x ) = x ^ { a x ^ \beta } $$ for some constant $ a \in \mathbb R $. The case of $ f ( x ) $ being constantly equal to $ 1 $ that you've found by yourself corresponds to the case $ a = 0 $.