Đáp án:

\( 9 f''(x) + [f'(x) - x]^2 = 9\)

\(\Rightarrow 9  f''(x) = -[f'(x) - x]^2. \)

\( \int \frac{f''(x) - 1}{[f'(x) - x]^2}  = -\frac{1}{9}  \Rightarrow  \int \frac{f''(x) - 1}{[f'(x) - x]^2} \, dx =  -\frac{1}{9} x + C. \)

\(\Rightarrow - \frac{1}{f'(x) - x} = -\frac{1}{9} x + C  \)

- \( f'(0) = 9 \implies C = -\frac{1}{9} \)

\( \frac{1}{f'(x) - x} = \frac{1}{9} (x + 1) \implies f'(x) = \frac{9}{x + 1} + x. \)

\( f(1) - f(0) = \int_0^1 f'(x) \, dx = \int_0^1 \left( x + \frac{9}{x+1} \right) \, dx. \)

\( = \frac{x^2}{2} + 9 \ln(x+1) \big|_0^1 = \frac{1}{2} + 9 \ln 2  \Rightarrow \boxed{C} \)