(a) We must have $e^{x}-3>0 \Leftrightarrow e^{x}>3 \Leftrightarrow x>\ln 3$. Thus, the domain of $f(x)=\ln \left(e^{x}-3\right)$ is $(\ln 3, \infty)$.
(b) $y=\ln \left(e^{x}-3\right) \Rightarrow e^{y}=e^{x}-3 \Rightarrow e^{x}=e^{y}+3 \Rightarrow x=\ln \left(e^{y}+3\right),$ so $f^{-1}(x)=\ln \left(e^{x}+3\right)$.
Now $e^{x}+3>0 \Rightarrow e^{x}>-3,$ which is true for any real $x,$ so the domain of $f^{-1}$ is $\mathbb{R}$.
