Is it possible to solve $f'(x)=f^{-1}(x)$? I am interested in proving what family of functions have the property
$$f'(x)=f^{-1}(x)$$
I've never dealt with a differential equation of this form, hence I could only go as far as to gather a little data:
$$f'(x)=f^{-1}(x)\implies f(f'(x))=x$$
$$\implies f''(x)f'(f'(x))=1$$
$$\implies f'(f'(x))=\frac{1}{f''(x)}$$
Let $f'=g$
$$\implies g(g(x))=\frac{1}{g'(x)}$$

Is this generally solvable?

Any and all information would be much useful.
 A: This is a partial answer.
In what follows we seek for a solution of the form $f(x)=ax^b$.
Then $f'(x)=abx^{b-1}$ while $f^{-1}(x)=(x/a)^{1/b}=a^{-1/b}x^{1/b}$.
So, setting, $f'(x)=f^{-1}(x)$, we obtain
$$
abx^{b-1}=a^{-1/b}x^{1/b},
$$
in which case $b-1=1/b$ equivalently $b^2-b-1=0$ or $b=\frac{1}{2}(1 \pm \sqrt{5})$,
while $ab=a^{-1/b}$ or $a^{b}b=1$ or choosing the positive value for $b$,
$$
a=b^{-1/b}=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}.
$$
Altogether,
$$
f(x)=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}} 
x^{\frac{1 + \sqrt{5}}{2}}
$$
possesses the required property: $f'=f^{-1}$.
Note that this function defines a global $C^1-$solution, if we define $x^b=|x|^{b-1}x$.
A: Well, we know that:
$$\text{y}\left(\text{y}^{-1}\left(x\right)\right)=x\tag1$$
Where $\text{y}^{-1}\left(x\right)$ is the inverse of the function $\text{y}\left(x\right)$.
So, using your problem we can write:
$$\text{y}'\left(x\right)=\text{y}^{-1}\left(x\right)\tag2$$
Note that a power of $x$ fits the bill for the differential equation, given in $(2)$. So, let's set:
$$\text{y}\left(x\right)=\text{A}x^\text{r}\tag3$$
We see that:
$$\text{y}'\left(x\right)=\text{r}\text{A}x^{\text{r}-1}\tag4$$
Now the inverse of the function is given by:
$$\text{y}^{-1}\left(x\right)=\left(\frac{x}{\text{A}}\right)^\frac{1}{\text{r}}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\tag5$$
So, we need to look at:
$$\text{r}\text{A}x^{\text{r}-1}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\space\Longleftrightarrow\space x^{\text{r}-1-\frac{1}{\text{r}}}=\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}\tag6$$

Now, to finish note that the RHS is a constant, so the LHS is a constant which means that $\text{r}-1-\frac{1}{\text{r}}=0$, which means that $\text{r}=\frac{1\pm\sqrt{5}}{2}$. So the LHS gives $x^0=1$, which means that $\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}=1$, and that gives $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$.

Concluding, we see that this is indeed true for $\text{r}=\frac{1+\sqrt{5}}{2}$ and $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$:
$$\text{y}\left(\text{y}'\left(x\right)\right)=\text{A}\left(\text{r}\text{A}x^{\text{r}-1}\right)^\text{r}=x\tag7$$
Note that when $\text{r}=\frac{1-\sqrt{5}}{2}$ there is no solution.
