Answer
a) $f\circ g =\dfrac{2}{2+3x}$
Domain: $\{x|x\ne-3,x\ne-\frac{3}{2},x\ne0 \}$
b) $g\circ f = \dfrac{2x+6}{x}$
Domain: $\{x|x\ne-3,x\ne0 \}$
c) $f\circ f = \dfrac{x}{4x+9}$
Domain: $\{x|x\ne-3, x=\ne-\frac{9}{4} \}$
d) $g\circ g = x$
Domain: $\{x|x\ne0\}$
Work Step by Step
a) $f\circ g = f(g(x)) = $
$\dfrac{\frac{2}{x}}{\frac{2}{x}+3}=$
$\dfrac{\frac{2}{x}}{\frac{2}{x}+\frac{3x}{x}}=$
$\dfrac{\frac{2}{x}}{\frac{2+3x}{x}}=$
$\dfrac{2(x)}{x(2+3x)}$
$\dfrac{2}{2+3x}$
Domain: $\{x|x\ne-3,x\ne-\frac{3}{2},x\ne0 \}$
b) $g\circ f = g(f(x)) =$
$\dfrac{2}{\frac{x}{x+3}}=$
$\dfrac{2(x+3)}{x}=$
$\dfrac{2x+6}{x}$
Domain: $\{x|x\ne-3,x\ne0 \}$
c) $f\circ f = f(f(x))=$
$\dfrac{\frac{x}{x+3}}{\frac{x}{x+3}+3}=$
$\dfrac{\frac{x}{x+3}}{\frac{x}{x+3}+\frac{3(x+3)}{x+3}}=$
$\dfrac{\frac{x}{x+3}}{\frac{x+3x+9}{x+3}}=$
$\dfrac{x(x+3)}{(x+3)(4x+9)}=$
$\dfrac{x}{4x+9}$
Domain: $\{x|x\ne-3, x=\ne-\frac{9}{4} \}$
d) $g\circ g = g(g(x))=$
$\dfrac{2}{\frac{2}{x}}=$
$\dfrac{2x}{2}=$
$x$
Domain: $\{x|x\ne0\}$