Find all $C^{1}$ functions $f: (0,+\infty) \to (0, +\infty)$ such that $f(x)^{f'(x)}=x$, $f(1)=1$. As the question title says, I'm trying to find all $C^1$ functions $f:(0, +\infty) \to (0, +\infty)$ which satisfy $f(x)^{f'(x)} = x$, and $f(1)=1$.
I know that $f(x)=x$ is one solution. When I put everything into the exponent, I get $f'(x) \ln{f(x)} = \ln{x}$, which gives me the implicit solution $f(x)(\ln{f(x)}-1) = x(\ln{x}-1)+C$, $C \in \mathbb{R}$. By inserting $(1,1)$ into the implicit solution, I get that the solution must satisfy $f(x)(\ln{f(x)}-1) = x(\ln{x}-1)$. The problem here is that I can't use Picard's theorem and claim uniqueness, because the expression $f'(x) = \frac{\ln{x}}{\ln{f(x)}}$ isn't defined for $(x, f(x))=(1,1)$.
Is there a different way to prove uniqueness, or is there another solution to this equation?

It seems that there is another solution to the equation, but for $0<x<e$, its value decreases from $e$ to $0$, and for $x>e$ the value becomes complex and is no longer real. This means that the derivative does not exist at that point. Other than that, it is a valid solution. Notice that, for $0<x<1$ the $W_0(\_)$ branch is used, and for $1<x<e$ the $W_{-1}(\_)$ branch is used in the formula given by Robert Israel. The key is that for any solution, $f'(1)^2=1$ and thus $f'(1)=1$ or else $f'(1)=-1$. The first leads to the easy $f(x):=x$ and the other to the solution using Lambert $W$. These are the only two possible solutions.
