To solve the equation

\[
|x + 3| + |x - 1| = \frac{16}{|y-2| + |y+2|},
\]

we can analyze each side separately.

### Step 1: Analyzing the left-hand side

The expression \( |x + 3| + |x - 1| \) can be simplified by considering the critical points where the expression changes: at \( x = -3 \) and \( x = 1 \). We can break it into intervals:

1. **Interval \( (-\infty, -3) \)**:
\[
|x + 3| + |x - 1| = -(x + 3) - (x - 1) = -2x - 2
\]

2. **Interval \( [-3, 1] \)**:
\[
|x + 3| + |x - 1| = (x + 3) - (x - 1) = 4
\]

3. **Interval \( (1, \infty) \)**:
\[
|x + 3| + |x - 1| = (x + 3) + (x - 1) = 2x + 2
\]

### Step 2: Analyze the right-hand side

The expression \( |y - 2| + |y + 2| \) also changes based on the critical points \( y = 2 \) and \( y = -2 \). Let's consider the intervals:

1. **Interval \( (-\infty, -2) \)**:
\[
|y - 2| + |y + 2| = -(y - 2) - (y + 2) = -2y
\]

2. **Interval \( [-2, 2] \)**:
\[
|y - 2| + |y + 2| = -(y - 2) + (y + 2) = 4
\]

3. **Interval \( (2, \infty) \)**:
\[
|y - 2| + |y + 2| = (y - 2) + (y + 2) = 2y
\]

### Step 3: Equate and solve

Now let's equate both sides across intervals:

1. **When \( x < -3 \)**:
\[
-2x - 2 = \frac{16}{-2y} \Rightarrow -2x - 2 = -\frac{8}{y} \Rightarrow y(-2x - 2) = -8 \Rightarrow y = \frac{8}{2x + 2}
\]

2. **When \( -3 \leq x \leq 1 \)**:
\[
4 = \frac{16}{4} \Rightarrow 4 = 4 \quad \text{(always true)}
\]

3. **When \( x > 1 \)**:
\[
2x + 2 = \frac{16}{2y} \Rightarrow 2y(2x + 2) = 16 \Rightarrow y = \frac{8}{x + 1}
\]

### Summary of Solutions

- For \( x < -3 \): \( y = \frac{8}{2x + 2} \)
- For \( -3 \leq x \leq 1 \): Any \( y \) satisfying \( |y-2| + |y+2| = 4 \).
- For \( x > 1 \): \( y = \frac{8}{x + 1} \)

You can evaluate specific values of \( x \) to find corresponding \( y \) values using these equations.