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?
 A: Well, here's an attempt using more basic machinery that avoids the Lambert W function.
I think all we need here is basic calculus.
Notice that for $x>1$, the relation $f'(x)\,\ln f(x)=\ln x$ implies that $f'(x)$ and $\ln f(x)$ have the same sign.
Similarly, for $x<1$ we must have $f'(x)$ and $\ln f(x)$ with opposite signs.
There are thus four possibilities
$\qquad(1.1)$: $f$ is increasing and $>1$ on $(1,+\infty)$, and $f$ is increasing and $<1$ on $(0,1)$.
$\qquad(1.2)$: $f$ is increasing and $>1$ on $(1,+\infty)$, and $f$ is decreasing and $>1$ on $(0,1)$.
$\qquad(2.1)$: $f$ is decreasing and $<1$ on $(1,+\infty)$, and $f$ is increasing and $<1$ on $(0,1)$.
$\qquad(2.2)$: $f$ is decreasing and $<1$ on $(1,+\infty)$, and $f$ is decreasing and $>1$ on $(0,1)$.
On the other hand, the relation $x(\ln x -1)=f(x)\, (\ln f(x) -1)$ implies that for $x\in(e,+\infty)$ we have $f(x)>e$.
This discards cases $(2.1)$ and $(2.2)$.

We know then that for $x>1$, any solution $f$ will be increasing and greater than one.
Now, let $f_1$ and $f_2$ be solutions to your differential equation and consider $g=f_1-f_2$. For $x>1$, we'll have
$$g'(x)=f_1'(x)-f_2'(x)=\frac{\ln x}{\ln f_1(x)}-\frac{\ln x}{\ln f_2(x)}=\frac{\ln x}{\ln f_1(x)\,\ln f_2(x)} \cdot \Big(\ln f_2(x) - \ln f_1(x)\Big)$$
It follows that for $x>1$, the sign of $g'(x)$ is that of $\ln f_2(x) - \ln f_1(x)$, which in turn is that of $f_2(x) - f_1(x) = -g(x)$.
Hence, for $x>1$ we have $g(x)g'(x)\leq0$.
This implies that $h(x)=g(x)^2$ satisfies $h'(x)\leq 0$, that is, $h$ is non increasing for $x>1$.
But letting $x\to 1$, we conclude $h\equiv 0$ throughout $(1,+\infty)$, and therefore $f_1(x)=f_2(x)$ for all $x\geq 1$.

Since we know that $f(x)=x$ is a solution on $[1,+\infty)$, we conclude it is the only solution on $[1,+\infty)$.
This is an important observation, because now case $(1.2)$ is discarded.
Since $f'\equiv1$ on $(1,+\infty)$, it would fail to be continuous at $x=1$ in case $(1.2)$, because then $f'<0$ on $(0,1)$.

It follows that the only possible case is $(1.1)$.
Now we may employ a similar argument to the one with $g$ up above to conclude that solutions are also unique for $x\leq 1$, and hence $f(x)=x$ is the only solution.
A: Letting $x = e t$ and $f(x) = e y$, the equation $f(x) (\ln f(x) - 1) = x (\ln x - 1)$ becomes $y \ln y = t \ln t$.  The solutions to this can be written as
$$y = \frac{t \ln t}{W(t \ln t)}$$
where $W$ is any branch of the Lambert W function. Now for $0 < t < 1$,
$-1/e < t \ln t < 0$, and in this interval there are two real branches of the Lambert $W$ function.
However, for $t > 1$, $t \ln t > 0$, and here there is only one real branch of $W$, the one that gives the solution $y = t$ or $f(x) = x$.  Thus there is only one $f$ if you insist on it being defined for $x \in (0, \infty)$.  There is another solution, but it is only defined for $x \in (0,e)$. 
A: 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.
