Đáp án:

Đặt \(h(x) = \frac{f'(x)}{[f(x)]^2}\).  

\(\Rightarrow \int_{4}^{8} h(x) dx = \int_{4}^{8} \frac{f'(x)}{[f(x)]^2} dx = -\frac{1}{f(x)} \big|_{4}^{8} = -2 + 4 = 2.\)  

\(\int_{4}^{8} [h(x)]^2 dx = 1.\)  

Tìm \(k\) sao cho: \(\int_{4}^{8} \left[h(x) - k\right]^2 dx = 0.\)  

\(\Leftrightarrow \int_{4}^{8} [h(x)]^2 dx - 2k \int_{4}^{8} h(x) dx + k^2 \int_{4}^{8} dx = 0\)  

\(\Leftrightarrow 1 - 4k + 4k^2 = 0 \iff (2k - 1)^2 = 0 \iff k = \frac{1}{2}.\)  

\(\Rightarrow h(x) =k \iff h(x) = \frac{1}{2} \iff \frac{f'(x)}{[f(x)]^2} = \frac{1}{2}\).  

\(\Rightarrow \frac{-1}{[f(x)]^2} =  \frac{1}{2} x + C.\)  

\(f(4) = \frac{1}{4} \Rightarrow  -4 = 2 + C \Rightarrow C = -6.\)  

\(\Rightarrow f(x) = \frac{-1}{\frac{x}{2}- 6} \Rightarrow f(6) = \frac{1}{3} \Rightarrow \boxed{D}.\)