Step 1: Identify the two conditions that restrict the domain.
The function is $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$. For this to be defined, the numerator logarithm must exist (argument strictly positive), and the denominator square root must be real and non-zero (expression strictly positive, since division by zero is not allowed). We handle each condition separately and then intersect the solution sets.
Step 2: Condition from the logarithm in the numerator.
For $\log_2(x+3)$ to be defined, its argument must be strictly positive: $x+3 > 0$, which gives $x > -3$. So the first condition contributes the interval $(-3, \infty)$.
Step 3: Condition from the square root in the denominator.
The denominator is $\sqrt{x^2+3x+2}$. For a square root in the denominator, the expression inside must be strictly greater than zero (equal to zero makes the fraction undefined): $x^2+3x+2 > 0$. We factor this quadratic next.
Step 4: Factor and solve the denominator inequality.
$x^2+3x+2 = (x+1)(x+2)$. The product is positive when both factors share the same sign. Case 1: both positive, i.e., $x > -1$. Case 2: both negative, i.e., $x < -2$. So the denominator condition gives $x \in (-\infty,-2) \cup (-1,\infty)$.
Step 5: Intersect both conditions to find the domain.
We need both conditions simultaneously. Condition 1: $x \in (-3,\infty)$. Condition 2: $x \in (-\infty,-2) \cup (-1,\infty)$. Their intersection is $(-3,-2) \cup (-1,\infty)$.
Step 6: Verify by checking test points.
Take $x = -2.5 \in (-3,-2)$: denominator $= (-1.5)(-0.5) = 0.75 > 0$ and $\log_2(0.5)$ is defined. Take $x = 0 \in (-1,\infty)$: denominator $= 2 > 0$, $\log_2(3)$ defined. At $x = -1$ or $x = -2$: denominator $= 0$, undefined.
\[ \boxed{(-3,-2)\cup(-1,\infty)} \]