\[\begin{array}{rcl}
f(x)&=&\frac{10^{2x}+1}{10^{2x}-1}\\
y(10^{2x}-1)&=&10^{2x}+1\\
y{10^{2x}}-y&=&10^{2x}+1\\
(y-1){10^{2x}}&=&y+1\\
{10^{2x}}&=&\frac{y+1}{y-1}\\
{{2x}}&=&log_{10}({\frac{y+1}{y-1}})\\
x&=&\frac{1}{2}log_{10}({\frac{y+1}{y-1}})\\
x&=&log_{10}(\sqrt{\frac{y+1}{y-1}})\\
\end{array}\]
ดังนั้น $f^{-1}(x)=log_{10}(\sqrt{\frac{x+1}{x-1}})$