Yet another attempt to $f'=f^{-1}$: via $(f^{-1})'=\frac 1{f'}$

Question

Here is another attempt to solve $f'=f^{-1}$, where $f\circ f^{-1}=x$. I know many solutions are available in different flavours, but I attempted a simple one based on the relation $(f^{-1})'=\frac 1{f'}$ (drop any constant of integration for simplicity): \begin{align}f'&=f^{-1}\\ f''&=\frac1{f'}\\ f'f''&=1\\ \frac{f'^2}2&=x \\ f'&=\sqrt{2x}\\ f(x)&=\frac13(2x)^{\frac32} \\ f^{-1}(x)&=\frac12 (3x)^{\frac23}.\end{align} Although the reasoning seems rigorous to me, I can see something is wrong, but not why. I also can't see why the application of $(f^{-1})'=\frac 1{f'}$ should cause any harm. What would you do?