If $\mathrm{f}\left(1\right)=3,{\mathrm{f}}^{\text{'}}\left(1\right)=2$ and ${\mathrm{f}}^{\text{'}\text{'}}\left(1\right)=4$ and ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)$, then $\mathrm{g}\text{'}\text{'}\left(3\right)$ is equal to

$$\begin{gathered} {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{x} \\ \Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{g}\right(\mathrm{x}\left)\right){\mathrm{g}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1\Rightarrow {\mathrm{g}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{g}\right(\mathrm{x}\left)\right)}\to \left(1\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow {\mathrm{g}}^{\text{'}\text{'}}\left(\mathrm{x}\right)=-\frac{{\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{g}\right(\mathrm{x}\left)\right){\mathrm{g}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)}{{\left({\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{g}\right(\mathrm{x}\left)\right)\right)}^2} \\ \Rightarrow {\mathrm{g}}^{\text{'}\text{'}}\left(\mathrm{x}\right)=-\frac{{\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{g}\right(\mathrm{x}\left)\right)}{{\left({\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{g}\right(\mathrm{x}\left)\right)\right)}^3} [\because \mathrm{From} (1\left)\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\mathrm{g}}^{\text{'}\text{'}}\left(3\right)=-\frac{{\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{g}\right(3\left)\right)}{{\left({\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{g}\right(3\left)\right)\right)}^3}\to \left(2\right) \\ \mathrm{let} \mathrm{g}\left(3\right)=\mathrm{x}\Rightarrow {\mathrm{f}}^{-1}\left(3\right)=\mathrm{x}\Rightarrow \mathrm{x}=1\Rightarrow \mathrm{g}\left(3\right)=1 \\ \mathrm{from} \left(2\right),\mathrm{ }{\mathrm{g}}^{\text{'}\text{'}}\left(3\right)=-\frac{{\mathrm{f}}^{\text{'}\text{'}}\left(1\right)}{{\left({\mathrm{f}}^{\mathrm{\text{'}}}\left(1\right)\right)}^3}=-\frac{4}{2^3}=-\frac{1}{2} \end{gathered}$$