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