I wanted to write the taylor series of the inverse of a bijective function around a point $y_0$ with the coefficients being in terms of the $\underline{\text{ derivatives of the function itself }}$ at the point $f(y=y_0)=x_0$, i.e: $$f^{-1}(y)=\sum_{n\geq 0}\left.\frac{d^nf^{-1}(y)}{dy^n}\right|_{y=y_0}\frac{(y-y_0)^n}{n!}\\ =\sum_{n\geq 0}o_n\left(\frac{dy}{df^{-1}(y)},\ldots,\frac{d^ny}{d(f^{-1}(y))^n}\right)_{f^{-1}(y=y_0)=x_0}\frac{(y-y_0)^n}{n!}$$ This helps me find the coefficients of the taylor series of an inverse function directly if I know the coefficients of the taylor series itself. And thus, I can invert a function easily.
Moving forward, I will use $y$ for the function and $x$ for its inverse and let $y_n$ be $\frac{d^ny}{dx^n}$ and the same for $x_n$.
We know: $$y_1=\frac{1}{x_1}$$ Differentiating both sides wrt $y$: $$\frac{dy_1}{dy}=-\frac{x_2}{x_1^2}$$ Then, multiplying both sides by $y_1$ and using the chain rule: $$\frac{dy_1}{dy}y_1=y_2=-\frac{x_2}{x_1^3}$$ You can imagine how deducing the $n$-th derivative this way will get easily chaotic. So, instead I decided to take the $n$-th derivative of the equation $y_1x_1=1$ wrt $y$ then look for a pattern. Taking the first derivative this way goes as follows: $$y_1x_1=1\\ (y_1)_yx_1+y_1x_2=0\\ y_2x_1+y_1^2x_2=0$$ The first four derivatives go as follows: $$y_1x_1=1\\ y_2x_1+y_1^2x_2=0\\ y_3x_1+3y_1y_2x_2+y_1^3x_3=0\\ y_4x_1+4y_1y_3x_2+3y_2^2x_2+3y_1^2y_2x_3+y_1^4x_4=0$$ I have this expression for the pattern that is easy to prove by induction: $$\frac{d^n}{dy^n}(x_1y_1)=\sum_{o_1p_1+o_2p_2=n}\left(n-\left|o_1-o_2\right|\right)y_{o_1}^{p_1}y_{o_2}^{p_2}x_{p_1+p_2}$$ However, I don't know how to use this to deduce the $n$-th derivative in terms of $x_k$'s only.