= Solution
Assume the needed <derivatives> exist continuously and $df/dg\ne0$ on a neighborhood, so the <inverse function theorem> provides a local inverse. Put $p=df/dg$, $q=d^2f/dg^2$, and $r=d^3f/dg^3$. The <chain rule> gives $d/df=p^{-1}d/dg$, so
$$
\frac{dg}{df}=\frac1p,\qquad
\frac{d^2g}{df^2}=\frac1p\frac d{dg}\frac1p=-\frac q{p^3}.
$$
Differentiate once more:
$$
\frac{d^3g}{df^3}=\frac1p\frac d{dg}(-qp^{-3})
=-\frac r{p^4}+\frac{3q^2}{p^5}.
$$
Thus the <first three derivatives of an inverse function> are
$$
\boxed{\frac{dg}{df}=\frac1{df/dg},\qquad
\frac{d^2g}{df^2}=-\frac{d^2f/dg^2}{(df/dg)^3},\qquad
\frac{d^3g}{df^3}=
\frac{3(d^2f/dg^2)^2-(df/dg)(d^3f/dg^3)}{(df/dg)^5}.}
$$
All forward <derivatives> on the right are evaluated at the corresponding value $g(f)$.
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