Đáp án:

Vì: \( \left( \frac{f'(x)}{f(x)} \right)' = \frac{f''(x) f(x) - (f'(x))^2}{(f(x))^2} \).

Nên: \( (f(x))^2 - f(x) \cdot f''(x) + (f'(x))^2 = 0 \).

 \( \Leftrightarrow\frac{f''(x) f(x) - (f'(x))^2}{(f(x))^2} = 1 \quad \Rightarrow \quad \left( \frac{f'(x)}{f(x)} \right)' = 1 \).

 \( \Rightarrow \frac{f'(x)}{f(x)} = x + c \quad \Rightarrow \quad \ln|{f(x)}| = \frac{x^2}{2} + cx + d \).

\( f(0) = 1 \Rightarrow d = 0 \).
\( f(2) = e^6 \Rightarrow 6 = 2 + 2c \quad \Rightarrow \quad c = 2 \).

\(\Rightarrow \ln{f(x)} = \frac{x^2}{2} + 2x \).

\(\Rightarrow \ln{f(1)} = \frac{1}{2} + 2 = \frac{5}{2} \quad \Rightarrow \quad f(1) = e^{\frac{5}{2}} \Rightarrow \boxed{\text{D}} \)