Đáp án:

Xét \( \left( \frac{f(x)}{x} \right)' = \frac{x f'(x) - f(x)}{x^2} \) : Khó xoay số

\( \left( \frac{f(x)}{x^2} \right)' = \frac{x^2 f'(x) - 2x f(x)}{x^4} \) : Có thể xoay số

Nhân 2 vế với \( x\):  
\( x^2 f(x) + 2x f(x) = x^2 f'(x) - x^4 \)

\( \Leftrightarrow x^2 f'(x) + x^4 = x^2 f'(x) - 2x f(x) \)

\( \Leftrightarrow \frac{x^2 f'(x) - 2x f(x)}{x^4} = \frac{f(x)}{x^2} + 1 \)

\( \Rightarrow \left( \frac{f(x)}{x^2} \right)' = \frac{f(x)}{x^2} + 1 \)

Đặt \( h(x) = \frac{f(x)}{x^2} + 1 \Rightarrow  h'(x) = h(x) \),  
\( \frac{h'(x)}{h(x)}= 1 \Rightarrow \ln h(x) = x + C \)  

\( f(1) = e \Rightarrow h(1) = e+1 \Rightarrow \ln(e+1) = 1 + C \),  
\( \Rightarrow C = \ln(e+1) - 1 \)

\( \Rightarrow |h(x)| = e^{x + \ln(e+1) - 1} \Rightarrow (e+1)e^{x-1} \)

\(\Rightarrow \frac{f(x)}{x^2} + 1 = (e+1)e^{x-1} \)

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

\(\Rightarrow f(2) = 4[(e+1)e - 1] = 4e^2 + 4e - 4 \Rightarrow \boxed{A}\)