If f(x) is defined on (0, 1), then the domain of definition of f(e^x) + f(ln|x|) is

Since the domain of f is (0, 1),

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<{\mathrm{e}}^{\mathrm{x}}<1\text{ and }0<\ln \left|\mathrm{x}\right|<1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\log 0<\mathrm{x}<\log 1\text{ and }{\mathrm{e}}^0<\left|\mathrm{x}\right|<{\mathrm{e}}^1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-\mathrm{\infty }<\mathrm{x}<0\text{ and }1<\left|\mathrm{x}\right|<\mathrm{e}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\in (-\mathrm{\infty },0)\text{ and }\mathrm{x}\in \left(\right(-\mathrm{\infty },-1)\cup (1,\mathrm{\infty }\left)\right)\cap (-\mathrm{e},\mathrm{e})\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\in (-\mathrm{\infty },0)\text{ and }\mathrm{x}\in (-\mathrm{e},-1)\cup (1,\mathrm{e})\Rightarrow \mathrm{x}\in (-\mathrm{e},-1)\end{array}$