Sử dụng phương pháp lấy nguyên hàm hai vế.Giải chi tiết:Ta có:\(\begin{array}{l}\,\,\,\,\,\,3{x^4}f\left( x \right) + {f^3}\left( x \right) = 2{x^5}f'\left( x \right)\\ \Leftrightarrow 3{x^4}f\left( x \right) - 2{x^5}f'\left( x \right) = - {f^3}\left( x \right)\\ \Leftrightarrow \dfrac{{3{x^2}f\left( x \right) - 2{x^3}f'\left( x \right)}}{{{f^3}\left( x \right)}} = - \dfrac{1}{{{x^2}}}\\ \Leftrightarrow \dfrac{{3{x^2}{f^2}\left( x \right) - 2{x^3}f'\left( x \right)f\left( x \right)}}{{{f^4}\left( x \right)}} = - \dfrac{1}{{{x^2}}}\\ \Leftrightarrow \dfrac{{\left( {{x^3}} \right)'{f^2}\left( x \right) - {x^3}\left[ {{f^2}\left( x \right)} \right]'}}{{{f^4}\left( x \right)}} = - \dfrac{1}{{{x^2}}}\\ \Leftrightarrow \left[ {\dfrac{{{x^3}}}{{{f^2}\left( x \right)}}} \right]' = - \dfrac{1}{{{x^2}}}\end{array}\)Lấy nguyên hàm hai vế ta được \(\dfrac{{{x^3}}}{{{f^2}\left( x \right)}} = \dfrac{1}{x} + C\).Mà \(f\left( 1 \right) = 1 \Rightarrow \dfrac{1}{1} = 1 + C \Leftrightarrow C = 0\), do đó \(\dfrac{{{x^3}}}{{{f^2}\left( x \right)}} = \dfrac{1}{x} \Leftrightarrow f\left( x \right) = {x^2}\).Vậy \(f\left( 3 \right) = 9\).Chọn D
