$f'(-1)=0\to -a+b=0$
$\to a=b$
$\to f'(x)=ax+\dfrac{a}{x^2}$
$\to f(x)=\dfrac{ax^2}{2}-\dfrac{a}{x}+C$ với $C$ là hằng số
$f(1)=4\to \dfrac{a}{2}-a+C=4$
$\to \dfrac{-a}{2}+C=4$
$f(-1)=2\to\dfrac{a}{2}+a+C=2$
$\to\dfrac{3a}{2}+C=2$
Giải hệ: $a=-1; C=\dfrac{7}{2}$
Vậy $f(x)=\dfrac{-1}{2}x^2+x+\dfrac{7}{2}$
$\to f(3)=2$