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?