$\begin{align}  & {{D}_{g}}:x\ne 1\,\,,\,\,{{D}_{f}}=x\ge 3 \\  & {{D}_{fog}}=\left\{ x\in {{D}_{g}}\left| g(x)\in {{D}_{f}} \right. \right\} \\  & \Rightarrow {{D}_{fog}}=\left\{ x\ne 1\left| \frac{x+1}{x-1}\ge 3 \right. \right\}\,\,\,(1) \\  & \frac{x+1}{x-1}\ge 3\Rightarrow \frac{x+1}{x-1}-3\ge 0\Rightarrow \frac{x+1-3x+3}{x-1}\ge 0 \\  & \Rightarrow \frac{-2x+4}{x-1}\ge 0\Rightarrow 1 \lt x\le 2\,\,\,(2) \\  & (1),(2)\Rightarrow {{D}_{fog}}=\left( 1,\left. 2 \right] \right. \\ \end{align}$