To solve the equation \(\frac{2x}{3} - 3 = \frac{3}{x + 3}\), follow these steps:

1. **Eliminate the fractions** by finding a common denominator. The common denominator for the fractions \(\frac{2x}{3}\) and \(\frac{3}{x+3}\) is \(3(x + 3)\).

2. **Multiply both sides of the equation** by the common denominator \(3(x + 3)\):

\[
3(x + 3) \left(\frac{2x}{3} - 3\right) = 3(x + 3) \left(\frac{3}{x + 3}\right)
\]

3. **Distribute** on both sides:

\[
(x + 3) \cdot 2x - 3 \cdot 3(x + 3) = 3 \cdot 3
\]

4. **Simplify** each term:

\[
2x(x + 3) - 9(x + 3) = 9
\]

5. **Expand** the terms:

\[
2x^2 + 6x - 9x - 27 = 9
\]

6. **Combine like terms**:

\[
2x^2 - 3x - 27 = 9
\]

7. **Move all terms to one side** to set the equation to zero:

\[
2x^2 - 3x - 27 - 9 = 0
\]

\[
2x^2 - 3x - 36 = 0
\]

8. **Solve the quadratic equation** \(2x^2 - 3x - 36 = 0\). You can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = -36\).

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-36)}}{2 \cdot 2}
\]

\[
x = \frac{3 \pm \sqrt{9 + 288}}{4}
\]

\[
x = \frac{3 \pm \sqrt{297}}{4}
\]

\[
x = \frac{3 \pm \sqrt{9 \cdot 33}}{4}
\]

\[
x = \frac{3 \pm 3\sqrt{33}}{4}
\]

\[
x = \frac{3(1 \pm \sqrt{33})}{4}
\]

So, the solutions are:

\[
x = \frac{3 + 3\sqrt{33}}{4} \quad \text{and} \quad x = \frac{3 - 3\sqrt{33}}{4}
\]

These are the two solutions to the equation \(\frac{2x}{3} - 3 = \frac{3}{x + 3}\).