1. The problem asks for the domain of the inverse function of $f(x) = \frac{2\sqrt{x+4}}{3}$.\n\n2. To find the domain of the inverse function, we first need to find the range of the original function $f(x)$, because the domain of the inverse is the range of the original.\n\n3. The function $f(x) = \frac{2\sqrt{x+4}}{3}$ involves a square root, so the expression inside the root must be non-negative: $$x+4 \geq 0$$\n\n4. Solving for $x$, we get: $$x \geq -4$$\n\n5. Since $\sqrt{x+4}$ is always non-negative, the smallest value of $f(x)$ occurs at $x = -4$: $$f(-4) = \frac{2\sqrt{-4+4}}{3} = \frac{2\cdot 0}{3} = 0$$\n\n6. As $x$ increases, $\sqrt{x+4}$ increases without bound, so $f(x)$ increases without bound. Therefore, the range of $f$ is: $$[0, \infty)$$\n\n7. The domain of the inverse function $f^{-1}$ is the range of $f$, so: $$\boxed{y \geq 0}$$\n\nThis means the inverse function is defined for all $y$ such that $y \geq 0$.