Multiply numerator and denominator by \(10^x\): \( y = \dfrac{10^{2x} - 1}{10^{2x} + 1} \).
Put \( t = 10^{2x} \), then \( y = \dfrac{t-1}{t+1} \). Solving, \( t = \dfrac{1+y}{1-y} \).
Hence, \( 10^{2x} = \dfrac{1+y}{1-y} \). Taking log base 10: \( 2x = \log_{10}\!\Big(\dfrac{1+y}{1-y}\Big) \).
The inverse function is:
\[
f^{-1}(y) = \tfrac{1}{2}\,\log_{10}\!\left(\dfrac{1+y}{1-y}\right), \quad |y|<1
\]
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