Answer
$f \circ g$: for all $D_{f\circ g}=\{x|x \in \mathbb{R}, x\ne 0\}$
$g \circ f$: for all $D_{g\circ f}=\{x|x \in \mathbb{R}, x \ne -1\}$,
$f \circ f$: for all $D_{f\circ f}=\{x|x \in \mathbb{R}, x\ne -\frac{1}{2}, x\ne -1\}$,
$g \circ g$: for all $D_{g\circ g}=\{x|x \in \mathbb{R}\}$
Work Step by Step
$f(x)=\frac{x}{x+1}$,
$D_f=\{x|x \in \mathbb{R}, x\ne -1 \}$
$g(x)=2x-1$,
$D_g=\{x|x \in \mathbb{R} \}$,
$f(g(x))=\frac{2x-1}{2x-1+1}=\frac{2x-1}{2x}$
$D_{f\circ g}=\{x|x \in \mathbb{R}, x\ne 0\}$
$g(f(x))=\frac{2x}{x+1}-1=\frac{x-1}{x+1}$
$D_{g\circ f}=\{x|x \in \mathbb{R}, x \ne -1\}$,
$f(f(x))=\frac{x}{2x+1}$
$D_{f\circ f}=\{x|x \in \mathbb{R}, x\ne -\frac{1}{2}, x\ne-1\}$,
$g(g(x))=2(2x-1)-1=4x-3$
$D_{g\circ g}=\{x|x \in \mathbb{R}\}$
