Let $f:R\to R$ and $g:R\to R$ be two functions defined by $f\left(x\right)={\log }_e\left(x^2+1\right)-e^{-x}+1$ and $g\left(x\right)=\frac{1-2e^{2x}}{e^x}$. Then, for which of the following range of $\alpha$, the inequality $f\left(g\left(\frac{(\alpha -1{)}^2}{3}\right)\right)>f\left(g\left(\alpha -\frac{5}{3}\right)\right)$ holds?

$$\begin{gathered} f\left(x\right)={\log }_e\left(x^2+1\right)-e^{-x}+1 \\ \Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{2\mathrm{x}}{{\mathrm{x}}^2+1}+{\mathrm{e}}^{-\mathrm{x}}>0 \forall \mathrm{x}\in \mathrm{R} \\ \Rightarrow \mathrm{f}\text{ is strictly increasing } \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{1-2{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}}={\mathrm{e}}^{-\mathrm{x}}-2{\mathrm{e}}^{\mathrm{x}} \\ \Rightarrow {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)=-\left(2{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\right)<0 \forall \mathrm{x}\in \mathrm{R} \\ \Rightarrow \mathrm{g}\text{ is decreasing } \\ \text{ Now }\mathrm{f}\left(\mathrm{g}\left(\frac{(\mathrm{\alpha }-1{)}^2}{3}\right)\right)>\mathrm{f}\left(\mathrm{g}\left(\mathrm{\alpha }-\frac{5}{3}\right)\right) \\ \Rightarrow \mathrm{g}\left(\frac{(\mathrm{\alpha }-1{)}^2}{3}\right)>\mathrm{g}\left(\mathrm{\alpha }-\frac{5}{3}\right) \\ \Rightarrow \frac{(\mathrm{\alpha }-1{)}^2}{3}<\mathrm{\alpha }-\frac{5}{3} \\ \Rightarrow {\mathrm{\alpha }}^2-5\mathrm{\alpha }+6<0 \\ \Rightarrow (\mathrm{\alpha }-2)(\mathrm{\alpha }-3)<0 \\ \Rightarrow \mathrm{\alpha }\in (2,3) \end{gathered}$$