Find the product of all values of $d$ such that $x^{3} +2x^{2} +3x +4 = 0$ and $x^{2} +dx +3 = 0$ have a common root.
1. Let \( \alpha \) be the common root of the polynomials \( x^3 + 2x^2 + 3x + 4 = 0 \) and \( x^2 + dx + 3 = 0 \).
2. Since \( \alpha \) is a root of \( x^2 + dx + 3 = 0 \), we have:
   \[
   \alpha^2 + d\alpha + 3 = 0 \quad \text{(1)}
   \]
3. Similarly, since \( \alpha \) is a root of \( x^3 + 2x^2 + 3x + 4 = 0 \), we have:
   \[
   \alpha^3 + 2\alpha^2 + 3\alpha + 4 = 0 \quad \text{(2)}
   \]
4. From equation (1), solve for \( \alpha^2 \):
   \[
   \alpha^2 = -d\alpha - 3
   \]
5. Substitute \( \alpha^2 = -d\alpha - 3 \) into equation (2):
   \[
   \alpha^3 + 2(-d\alpha - 3) + 3\alpha + 4 = 0
   \]
   Simplify:
   \[
   \alpha^3 - 2d\alpha - 6 + 3\alpha + 4 = 0
   \]
   \[
   \alpha^3 + (3 - 2d)\alpha - 2 = 0 \quad \text{(3)}
   \]
6. From equation (1), multiply both sides by \( \alpha \):
   \[
   \alpha^3 + d\alpha^2 + 3\alpha = 0
   \]
   Substitute \( \alpha^2 = -d\alpha - 3 \):
   \[
   \alpha^3 + d(-d\alpha - 3) + 3\alpha = 0
   \]
   Simplify:
   \[
   \alpha^3 - d^2\alpha - 3d + 3\alpha = 0
   \]
   \[
   \alpha^3 + (3 - d^2)\alpha - 3d = 0 \quad \text{(4)}
   \]
7. Equate the coefficients of \( \alpha \) from equations (3) and (4):
   \[
   3 - 2d = 3 - d^2
   \]
   Simplify:
   \[
   -2d = -d^2
   \]
   \[
   d^2 - 2d = 0
   \]
   Factorize:
   \[
   d(d - 2) = 0
   \]
   Thus, \( d = 0 \) or \( d = 2 \).

8. The product of all values of \( d \) is:
   \[
   0 \times 2 = 0
   \]

The final answer is \(\boxed{0}\).