= First three derivatives of an inverse function
{title2=$g'=1/f',\quad g''=-f''/(f')^3,\quad g'''=[3(f'')^2-f'f''']/(f')^5$}
Let a three-times <continuously differentiable> real <function> $f(g)$ have $df/dg\ne0$. The <inverse function theorem> gives a local inverse $g(f)$. Writing $p=df/dg$, $q=d^2f/dg^2$ and $r=d^3f/dg^3$, its <derivatives> are
$$
\frac{dg}{df}=\frac1p,\qquad
\frac{d^2g}{df^2}=-\frac q{p^3},\qquad
\frac{d^3g}{df^3}=\frac{3q^2-pr}{p^5}.
$$
Indeed $d/df=p^{-1}d/dg$ by the <chain rule>. Apply this operator successively to $g$, $p^{-1}$ and $-qp^{-3}$. Every forward derivative on the right is evaluated at the inverse image $g(f)$.
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