Can a map $\phi$ be defined such that $\phi\left(\frac{f\left(x\right)f''\left(x\right)}{f'\left(x\right)^{2}}\right)=f\left(x\right)$?

Question

As per the title, does a mapping $\phi:\mathbb{R}\rightarrow\mathbb{R}$ exist such that

$$\phi\left(\frac{f\left(x\right)f''\left(x\right)}{f'\left(x\right)^{2}}\right)=f\left(x\right)\text{?}$$

For clarification, if we defined $H(f'(x))=f(x)$, then $H$ is the operator $\int_0^x f'(s)ds=f(x)$. Note that this is equivalent to solving the second-order nonlinear ODE for $f(x)$

$$\frac{f\left(x\right)f''\left(x\right)}{f'\left(x\right)^{2}}=g(x).$$

Any help is much appreciated.

Answer 1

As mentioned in the comment, I think that the framing in terms of a mapping/operator $\phi$ is a little broken, but for suitable $f, g$, we can solve the differential equation.

Put $h(x) = f(x)/f'(x)$. Then $$ h'(x) = \frac{-f'(x)^2 + f(x)f''(x)}{f'(x)^2} = \frac{f(x)f''(x)}{f'(x)^2} - 1. $$ So the differential equation is then $$ h'(x) + 1 = g(x) $$ which we can solve for $h$ given an antiderivative for $g - 1$. We can recover $f$ from $h$ using $h(x) = \frac{1}{\frac{d}{dx} \log(f(x))}$.