$\text{ The range of the function }f\left(x\right)=\frac{e^x-e^{\left|x\right|}}{e^x+e^{\left|x\right|}}\text{ is }$

$f\left(x\right)=\frac{e^x-e^{\left|x\right|}}{e^x+e^{\left|x\right|}}=\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\ge 0\\ \frac{e^x-e^{-x}}{e^x+e^{-x}},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x<0\end{array}\right.$

$\text{ Clearly, }f\left(x\right)\text{ is identically zero if }x\ge 0----\left(1\right)$

$\begin{array}{l}\because \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x<0\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{e}^{2x}<1\text{ or }0<e^{2x}<1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<\frac{1+y}{1-y}<1\end{array}$

$\begin{array}{l}\text{ or }\frac{1+y}{1-y}>0\text{ and }\frac{1+y}{1-y}<1\\ \text{ or }(y+1)(y-1)<0\text{ and }\frac{2y}{1-y}<0\end{array}$

$\text{ i.e., }-1<y<1\text{ and }y<0\text{ or }y>1$

$\text{ or }-1<y<0------\left(2\right)$

$\text{ Combining (1) and (2), we get }-1<y\le 0\text{ or Range }=(-1,0]\text{ . }$