Let $ f : \mathbf {R} \to \mathbf {R} $ be a smooth function such that $ f'(x)^2 = f(x) f''(x) $ for all $x$. Suppose $f(0)=1$ and $f^{(4)} (0) = 9$. Find all possible values of $f'(0)$.
1. Let \( f : \mathbf{R} \to \mathbf{R} \) be a smooth function such that \( f'(x)^2 = f(x) f''(x) \) for all \( x \). We are given \( f(0) = 1 \) and \( f^{(4)}(0) = 9 \). We need to find all possible values of \( f'(0) \).

2. Define \( f'(0) = a \), \( f''(0) = b \), and \( f'''(0) = c \).

3. Substituting \( x = 0 \) into the given equation \( f'(x)^2 = f(x) f''(x) \), we get:
   \[
   a^2 = f(0) f''(0) \implies a^2 = 1 \cdot b \implies a^2 = b
   \]

4. Differentiating both sides of the original equation \( f'(x)^2 = f(x) f''(x) \) with respect to \( x \), we get:
   \[
   2 f'(x) f''(x) = f'(x) f''(x) + f(x) f'''(x)
   \]
   Simplifying, we obtain:
   \[
   f'(x) f''(x) = f(x) f'''(x)
   \]

5. Substituting \( x = 0 \) into the differentiated equation, we get:
   \[
   a b = 1 \cdot c \implies ab = c
   \]

6. Differentiating the equation \( f'(x) f''(x) = f(x) f'''(x) \) again with respect to \( x \), we get:
   \[
   f''(x) f''(x) + f'(x) f'''(x) = f'(x) f'''(x) + f(x) f^{(4)}(x)
   \]
   Simplifying, we obtain:
   \[
   (f''(x))^2 = f(x) f^{(4)}(x)
   \]

7. Substituting \( x = 0 \) into the twice-differentiated equation, we get:
   \[
   b^2 = 1 \cdot f^{(4)}(0) \implies b^2 = 9 \implies b = \pm 3
   \]

8. From \( a^2 = b \), we have:
   \[
   a^2 = 3 \implies a = \pm \sqrt{3}
   \]

9. We check for consistency with the second equation \( ab = c \). If \( b = 3 \), then \( c = 3a \). If \( b = -3 \), then \( c = -3a \). Both cases are consistent with the given conditions.

Conclusion:
\[
a = \boxed{\pm \sqrt{3}}
\]