Differential equations guarantees for function inverse

Suppose we had an invertible function $f:\mathbb{R} \to \mathbb{R}$, and suppose further we knew there was a differential equation that governed it such that $N\left[f\right] = 0$. A natural question to ask is "Does this mean a differential equation also governs the inverse?". Solving this is pretty straightforward. If $f(x) = y$ and $f^{-1}(y) = x$ we know that:$$\left[f^{-1}\right]'(y) = \frac{1}{f'(f^{-1}(y))}$$From here we invert both sides to define $f'(f^{-1}(y))$ as a function of $\left[f^{-1}\right]'(y)$:$$\frac{1}{\left[f^{-1}\right]'(y)} = f'(f^{-1}(y))$$From simple chain rule we know that $\left(f^{(n)}(f^{-1}(y))\right)' = f^{(n+1)}(f^{-1}(y))\left[f^{-1}\right]'(y)$. Thus we can say:$$\left(\frac{1}{\left[f^{-1}\right]'(y)}\right)^{(n)} = f^{(n+1)}(f^{-1}(y))\left[f^{-1}\right]'(y)$$And by rearranging:$$\frac{1}{\left[f^{-1}\right]'(y)}\left(\frac{1}{\left[f^{-1}\right]'(y)}\right)^{(n)} = f^{(n+1)}(f^{-1}(y))$$This gives us the means to generate any derivative of $f$ as a function of the derivatives of the inverse. Particularly, as the $\left(\frac{1}{\left[f^{-1}\right]'(y)}\right)^{(n)}$ term will always be a ratio of polynomials formed from powers of derivatives of the inverse, that means that if $N[f]=0$ is a polynomial of derivatives of $f$, then so will the differential equation that governs $f^{-1}$. Furthermore, it will be of the same order, giving us a certain 'niceness'. Note as well that $N[f]=0$ holds regardless of the path it is evaluated along.

N.B. When I talk about a 'polynomial of derivatives', I mean a combination of terms of the form $\prod_{k=0} \left(f^{(k)}\left(x\right)\right)^{\alpha_k}$, so something like $\left(f\right)^3\left(f'\right)^2-3.2\left(f''\right)+7f^9$