Given,

$f:\left(0,1\right)\to \mathrm{ℝ}$ be a function defined by $f\left(x\right)=\frac{1}{1-e^{-x}}$, and $g\left(x\right)=\left(f\left(-x\right)-f\left(x\right)\right)$

Now using $g\left(x\right)=f\left(-x\right)-f\left(x\right)$ we get,

$g\left(x\right)=\frac{1}{1-e^x}-\frac{e^x}{e^x-1}$

$\Rightarrow g\left(x\right)=\frac{1+e^x}{1-e^x}$

Now differentiating the above function with respect to $x$ we get,

$\Rightarrow g^{\text{'}}\left(x\right)=\frac{2e^x}{{\left(1-e^x\right)}^2}>0$

Since $g^{\text{'}}\left(x\right)$ is always positive

So,$g\left(x\right)$ is one-one in $\left(0,1\right)$

And $g\left(x\right)$ is increasing in $\left(0,1\right)$