1. Given that $\tan \alpha = \frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, we need to find $\tan \beta$ such that $\tan 2\beta = \tan 3\alpha$.

2. First, we use the triple angle formula for tangent:
$$\tan 3\alpha = \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3\tan^2 \alpha}$$
Substituting $\tan \alpha = \frac{p}{q}$, we get:
$$\tan 3\alpha = \frac{3 \cdot \frac{p}{q} - \left(\frac{p}{q}\right)^3}{1 - 3 \left(\frac{p}{q}\right)^2} = \frac{\frac{3p}{q} - \frac{p^3}{q^3}}{1 - \frac{3p^2}{q^2}} = \frac{\frac{3pq^2 - p^3}{q^3}}{\frac{q^2 - 3p^2}{q^2}} = \frac{p(3q^2 - p^2)}{q(q^2 - 3p^2)}$$

3. Let $A = \frac{q(q^2 - 3p^2)}{p(3q^2 - p^2)}$. Then:
$$\tan 3\alpha = \frac{1}{A}$$

4. According to the problem, $\tan 2\beta = \tan 3\alpha = \frac{1}{A}$. Using the double angle formula for tangent:
$$\tan 2\beta = \frac{2\tan \beta}{1 - \tan^2 \beta}$$
Setting this equal to $\frac{1}{A}$, we get:
$$\frac{2\tan \beta}{1 - \tan^2 \beta} = \frac{1}{A}$$

5. Cross-multiplying and rearranging terms, we obtain:
$$2A \tan \beta = 1 - \tan^2 \beta \implies \tan^2 \beta + 2A \tan \beta - 1 = 0$$

6. Solving this quadratic equation for $\tan \beta$, we use the quadratic formula:
$$\tan \beta = \frac{-2A \pm \sqrt{(2A)^2 + 4}}{2} = -A \pm \sqrt{A^2 + 1}$$

7. For $\tan \beta$ to be rational, $\sqrt{A^2 + 1}$ must be rational. This implies that $A^2 + 1$ must be a perfect square. Let $A^2 + 1 = k^2$ for some integer $k$. Then:
$$k^2 - A^2 = 1 \implies (k - A)(k + A) = 1$$

8. We now express $A$ in terms of $p$ and $q$:
$$A = \frac{q(q^2 - 3p^2)}{p(3q^2 - p^2)}$$

9. We need to show that $A^2 + 1$ is a perfect square if and only if $p^2 + q^2$ is a perfect square. We compute $A^2 + 1$:
$$A^2 + 1 = \left(\frac{q(q^2 - 3p^2)}{p(3q^2 - p^2)}\right)^2 + 1 = \frac{q^2(q^2 - 3p^2)^2 + p^2(3q^2 - p^2)^2}{p^2(3q^2 - p^2)^2}$$

10. Simplifying the numerator:
$$q^2(q^2 - 3p^2)^2 + p^2(3q^2 - p^2)^2 = q^2(q^4 - 6q^2p^2 + 9p^4) + p^2(9q^4 - 6q^2p^2 + p^4)$$
$$= q^6 - 6q^4p^2 + 9q^2p^4 + 9p^2q^4 - 6p^4q^2 + p^6 = q^6 + p^6 + 3q^4p^2 + 3p^4q^2$$
$$= (q^2 + p^2)^3$$

11. Therefore:
$$A^2 + 1 = \frac{(q^2 + p^2)^3}{p^2(3q^2 - p^2)^2}$$

12. For $A^2 + 1$ to be a perfect square, $(q^2 + p^2)^3$ must be a perfect square, which implies $q^2 + p^2$ must be a perfect square.

Thus, $\tan \beta$ is rational if and only if $p^2 + q^2$ is the square of an integer.

$\blacksquare$