1. **Rearrange the equation:**

  Divide both sides by (2xy + x²) and (3y² + 2xy) to separate the variables:

  (dy/dx) / (3y² + 2xy) = 1 / (2xy + x²)

2. **Simplify:**

  Factor out y and x from the denominators:

  (dy/dx) / [y(3y + 2x)] = 1 / [x(2y + x)]

3. **Integrate both sides:**

  ∫ (1/[y(3y + 2x)]) dy = ∫ (1/[x(2y + x)]) dx

  This requires partial fractions. Let's focus on the left side first:

  1/[y(3y + 2x)] = A/y + B/(3y + 2x)

  Solving for A and B, we get A = 1/(2x) and B = -3/(2x).

  Now we can integrate:

  ∫ (1/(2x)) (1/y) dy - ∫ (3/(2x)) (1/(3y + 2x)) dy = ∫ (1/[x(2y + x)]) dx

4. **Integrate the terms:**

  (1/(2x)) ln|y| - (3/(2x)) (1/3) ln|3y + 2x| = ∫ (1/[x(2y + x)]) dx

5. **Simplify and integrate the right side:**

  (1/(2x)) ln|y| - (1/(2x)) ln|3y + 2x| = ∫ (1/[x(2y + x)]) dx

  The right side also requires partial fractions:

  1/[x(2y + x)] = C/x + D/(2y + x)

  Solving for C and D, we get C = 1/(2y) and D = -1/(2y).

  Now we can integrate:

  (1/(2x)) ln|y| - (1/(2x)) ln|3y + 2x| = (1/(2y)) ln|x| - (1/(2y)) ln|2y + x| + K (where K is the constant of integration)

6. **Combine terms:**

  (1/(2x)) ln|y/(3y + 2x)| = (1/(2y)) ln|x/(2y + x)| + K

7. **Solve for y (optional):**

  This equation implicitly defines y as a function of x. It's difficult to explicitly solve for y. However, you can leave the solution in this implicit form. 

**Final implicit solution:** 

(1/(2x)) ln|y/(3y + 2x)| = (1/(2y)) ln|x/(2y + x)| + K