The expression under the square root must be non-negative.

For $\sqrt{x+7}$ to be defined, we must have:

$$x+7 \ge 0$$ $$x \ge -7$$

For the fraction to be defined, the denominator $x^2-4$ cannot be equal to zero:

$$x^2-4 \ne 0$$

We can factor the denominator as a difference of squares:

$$(x-2)(x+2) \ne 0$$

This means that:

$$x-2 \ne 0 \implies x \ne 2$$ $$x+2 \ne 0 \implies x \ne -2$$

We need to satisfy both $x \ge -7$ and $x \ne 2$ and $x \ne -2$.
The interval $x \ge -7$ is $[-7, \infty)$.
From this interval, we must exclude the values $x=2$ and $x=-2$. Both of these values are within the interval $[-7, \infty)$.
So, the domain of $f(x)$ is all $x$ such that $x \ge -7$ but $x \ne -2$ and $x \ne 2$.
In interval notation, this is:

$$[-7, -2) \cup (-2, 2) \cup (2, \infty)$$

The domain of $f(x)$ is $\boxed{[-7, -2) \cup (-2, 2) \cup (2, \infty)}$.