1. Given functions: $f(x) = \sqrt{x + 3}$ and $g(x) = \frac{1}{x}$.\n2. Find $g(f(x))$: Substitute $f(x)$ into $g(x)$ to get $g(f(x)) = \frac{1}{f(x)} = \frac{1}{\sqrt{x + 3}}$.\n3. Determine the domain of $g(f(x))$: The expression inside the square root must be non-negative, so $x + 3 \geq 0 \Rightarrow x \geq -3$.\n4. Also, the denominator cannot be zero, so $\sqrt{x + 3} \neq 0 \Rightarrow x + 3 \neq 0 \Rightarrow x \neq -3$.\n5. Combining these, the domain is $x > -3$.\n6. Final answer: $g(f(x)) = \frac{1}{\sqrt{x + 3}}$ with domain $x > -3$.