Answer
$$(f \circ g)(x)=\sqrt{\sqrt{x^2+3}-3}$$
$$(g \circ f)(x)=\sqrt{x}$$
$$D_{f \circ g}= (- \infty , \sqrt{6}] \cup [\sqrt{6} , + \infty )$$
$$D_{g \circ f}= [3, + \infty )$$
Work Step by Step
$$(f \circ g)(x)=f(g(x))=\sqrt{\sqrt{x^2+3}-3}$$
$$(g \circ f)(x)=g(f(x))=\sqrt{(\sqrt{x-3})^2+3}=\sqrt{x}$$
The domain of $f \circ g$ consists of all $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$. So we have
$$D_{f \circ g}= \{ x \in \mathbb{R} \mid \sqrt{x^2+3} \ge 3 \}=(- \infty , \sqrt{6}] \cup [\sqrt{6} , + \infty )$$ $$D_{g \circ f}= \{ x \ge 3 \mid \sqrt{x-3} \in \mathbb{R} \}=[3, + \infty ).$$
