A formula for n-derivative of the inverse of a function? Let $y=f^{-1}(x)$. As we know:
\begin{align}  \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{{f}'(y)}  \end{align}
Thereof we have:
\begin{align}  \frac{\mathrm{d^2} y}{\mathrm{d} x^2}=\frac{-{f}''(y)}{({f}'(y))^3}  \end{align}
\begin{align}  \frac{\mathrm{d^3} y}{\mathrm{d} x^3}=\frac{3({f}''(y))^2-{f}'(y){f}'''(y)}{({f}'(y))^5}  \end{align}
Is there a general rule for
\begin{align}  \frac{\mathrm{d^n} y}{\mathrm{d} x^n}=?\end{align}

One expression is 
$$y^{(n)}(x_0)=\bigl((y-y_0)(\frac{y-y_0}{f(y)-x_0})^{n+1} f'(y)\bigr)^{(n)}|_{y=y_0}$$
where $y_0=f^{-1}(x_0)$ (it is a constant). Supposing $f$ is holomorphic, it can be derived from the residue theorem:
$y^{(n)}(x_0)=n!/2\pi i\,\oint (y-y_0)/(x-x_0)^{n+1} dx =n!/2\pi i\,\oint  ((y-y_0)/(x-x_0))^{n+1}/(y-y_0)^{n}f'\,dy$ 
(the residue theorem is taught as a way of computing integrals in terms of derivatives, but in some cases it is a means of computing derivatives in terms of integrals)
BTW, it becomes more entertaining when $h$ is another function and you want to find the n-th derivative of $h(y(x))$. Supposing $f(0)=0$, and that we compute at $0$ (i.e. $x_0=y_0=0$, we get
$$\bigl(h(y)(\frac{y}{f(y)})^{n+1} f'(y)\bigr)^{(n)}|_{y=0}$$
As an example, if $h(y)=1/(1-y)$, $x=ye^{-y}$, we get the Taylor series
$$h(y)=\sum_n\frac{n^n}{n!}x^n.$$
