We just have to write one function inside the other:
[tex](f\circ g)(x) = f(g(x))\\\\(f\circ g)(x) = f\left(\dfrac{13}{x+3}\right)\\\\(f\circ g)(x) = \dfrac{2}{\left(\dfrac{13}{x+3}\right)+3}\\\\(f\circ g)(x) = \dfrac{2}{~~\dfrac{13+3(x+3)}{x+3}~~}\\\\(f\circ g)(x) = \dfrac{2(x+3)}{13+3(x+3)}\\\\\boxed{(f\circ g)(x) = \dfrac{2x+6}{3x+22}}[/tex]
Since we have a fraction with x in the denominator, we have to remove from the domain the value of that makes null the denominator. Thus,
[tex]3x+22\neq0\Longrightarrow 3x\neq22\Longrightarrow x\neq \dfrac{22}{3}[/tex]
Furthermore, we also have to remove the value of x that annuls the denominator of g(x), because that's the first function that we use. So:
[tex]x+3\neq0\Longrightarrow x\neq-3[/tex]
Hence, the domain is:
[tex]D((f\circ g)(x)) = \left]-\infty,-3\right[\cup \left]-3,\dfrac{22}{3}\right[\cup\left]\dfrac{22}{3},\infty\right[[/tex]
