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

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

Clearly, f(x) is identically zero if $x\ge 0$                       (1)

If $x<0, \mathrm{let} y=f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \mathrm{or} e^{2x}=\frac{1+y}{1-y}$

$\begin{array}{l} x<0\\ e^{2x}<1 \mathrm{or} 0<e^{2x}<1\\ \therefore 0<\frac{1+y}{1-y}<1\\ \mathrm{or} \frac{1+y}{1-y}>0 \mathrm{and} \frac{1+y}{1-y}<1\\ \mathrm{or} \left(y+1\right)\left(y-1\right)<0 \mathrm{and} \frac{2y}{1-y}<0\\ \mathrm{i}.\mathrm{e}., -1<y<1 \mathrm{and} y<0 \mathrm{or} y>1\end{array}$

$\mathrm{or} -1<y<0$                                                (2)

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