Let $ X$ be the set of all positive integers greater than or equal to $ 8$ and let $ f: X\rightarrow X$ be a function such that $ f(x\plus{}y)\equal{}f(xy)$ for all $ x\ge 4, y\ge 4 .$ if $ f(8)\equal{}9$, determine $ f(9) .$
1. Given the function \( f: X \rightarrow X \) where \( X \) is the set of all positive integers greater than or equal to 8, and the functional equation \( f(x+y) = f(xy) \) for all \( x \geq 4 \) and \( y \geq 4 \), we need to determine \( f(9) \) given that \( f(8) = 9 \).

2. Let's start by analyzing the given functional equation \( f(x+y) = f(xy) \). We know that \( f(8) = 9 \). We need to find \( f(9) \).

3. Consider \( x = 4 \) and \( y = 4 \):
   \[
   f(4+4) = f(4 \cdot 4) \implies f(8) = f(16)
   \]
   Since \( f(8) = 9 \), it follows that:
   \[
   f(16) = 9
   \]

4. Next, consider \( x = 4 \) and \( y = 5 \):
   \[
   f(4+5) = f(4 \cdot 5) \implies f(9) = f(20)
   \]
   We need to find \( f(20) \) to determine \( f(9) \).

5. Consider \( x = 4 \) and \( y = 6 \):
   \[
   f(4+6) = f(4 \cdot 6) \implies f(10) = f(24)
   \]

6. Consider \( x = 4 \) and \( y = 7 \):
   \[
   f(4+7) = f(4 \cdot 7) \implies f(11) = f(28)
   \]

7. Consider \( x = 4 \) and \( y = 8 \):
   \[
   f(4+8) = f(4 \cdot 8) \implies f(12) = f(32)
   \]

8. Notice that we can generalize this pattern. For any \( x \geq 4 \) and \( y \geq 4 \), we have:
   \[
   f(x+y) = f(xy)
   \]
   This implies that the function \( f \) is constant for all \( x \geq 8 \).

9. Since \( f(8) = 9 \) and \( f \) is constant for all \( x \geq 8 \), it follows that:
   \[
   f(9) = 9
   \]

The final answer is \( \boxed{9} \).