To solve the equation

\[
|3x + 1| + |3x - 5| = \frac{12}{(y + 3)^2} + 2,
\]

we need to break it down into parts:

### Step 1: Set Up Cases for the Absolute Values

The expressions \( |3x + 1| \) and \( |3x - 5| \) change based on the value of \( x \).

1. **Case 1**: \( x < -\frac{1}{3} \)
- Here, \( |3x + 1| = -(3x + 1) = -3x - 1 \)
- \( |3x - 5| = -(3x - 5) = -3x + 5 \)
- So the equation becomes:
\[
-3x - 1 - 3x + 5 = \frac{12}{(y + 3)^2} + 2
\]

2. **Case 2**: \( -\frac{1}{3} \leq x < \frac{5}{3} \)
- Here, \( |3x + 1| = 3x + 1 \)
- \( |3x - 5| = -(3x - 5) = -3x + 5 \)
- The equation becomes:
\[
3x + 1 - 3x + 5 = \frac{12}{(y + 3)^2} + 2
\]

3. **Case 3**: \( x \geq \frac{5}{3} \)
- Here, \( |3x + 1| = 3x + 1 \)
- \( |3x - 5| = 3x - 5 \)
- The equation becomes:
\[
3x + 1 + 3x - 5 = \frac{12}{(y + 3)^2} + 2
\]

### Step 2: Solve Each Case

Now let's solve each case:

#### Case 1:
\[
-6x + 4 = \frac{12}{(y + 3)^2} + 2.
\]
This simplifies to:
\[
-6x + 2 = \frac{12}{(y + 3)^2}.
\]

#### Case 2:
\[
6 = \frac{12}{(y + 3)^2} + 2 \implies 4 = \frac{12}{(y + 3)^2} \implies (y + 3)^2 = 3.
\]

#### Case 3:
\[
6x - 4 = \frac{12}{(y + 3)^2} + 2 \implies 6x - 6 = \frac{12}{(y + 3)^2}.
\]

### Step 3: Find Solutions for \( y \)

1. **Case 1 and Case 3** can be solved based on specific values of \( x \).

2. **Case 2** gives us:
\[
y + 3 = \pm \sqrt{3} \implies y = -3 \pm \sqrt{3}.
\]

### Final Answer

The potential solutions for \( y \) will thus depend on the values of \( x \), and you will need to consider the constraints you've put in place for each case to find specific \( x, y \) pairs that satisfy the original equation.

After solving for all cases:

- For Case 2 in particular, we find the values of \( y \) as \( y = -3 \pm \sqrt{3} \).

Do further calculations to analyze \( x \) values in connection to \( y \).