Answer
a) $f\circ g =\dfrac{x}{-2+3x}$
Domain: $\{x|x\ne-3,x\ne0,x\ne\frac{2}{3} \}$
b) $g\circ f = -2x-6$
Domain: $\{x|x\ne-3,x\ne0 \}$
c) $f\circ f =\dfrac{x+3}{3x+4}$
Domain: $\{x|x\ne-3,x\ne-\frac{4}{3} \}$
d) $g\circ g = x$
Domain: $\{x|x\ne0\}$
Work Step by Step
a) $f\circ g = f(g(x)) = $
$\dfrac{1}{-\frac{2}{x}+3}=$
$\dfrac{1}{\frac{-2}{x}+\frac{3x}{x}}=$
$\dfrac{1}{\frac{-2+3x}{x}}=$
$\dfrac{x}{-2+3x}$
Domain: $\{x|x\ne-3,x\ne0,x\ne\frac{2}{3} \}$
b) $g\circ f = g(f(x)) =$
$-\dfrac{2}{\frac{1}{x+3}}=$
$-\dfrac{2(x+3)}{1}=$
$-2x-6$
Domain: $\{x|x\ne-3,x\ne0 \}$
c) $f\circ f = f(f(x))=$
$\dfrac{1}{\frac{1}{x+3}+3}=$
$\dfrac{1}{\frac{1}{x+3}+\frac{3(x+3)}{x+3}}=$
$\dfrac{1}{\frac{1+3x+3}{x+3}}=$
$\dfrac{1(x+3)}{3x+4}=$
$\dfrac{x+3}{3x+4}$
Domain: $\{x|x\ne-3,x\ne-\frac{4}{3} \}$
d) $g\circ g = g(g(x))=$
$-\dfrac{2}{-\frac{2}{x}}=$
$-\dfrac{2x}{-2}=$
$x$
Domain: $\{x|x\ne0\}$