[tex]\dfrac{3x+2}{(x+3)^2}=\dfrac{A}{x+3}+\dfrac{B}{(x+3)^2}\\\\\\3x+2=A(x+3)+B\\\\3x+2=Ax+3A+B\\\\\left \{\begin{array}{l} {{3=A} \\ {2=3A+B}} \end{array}\right. \\\\\\\left \{\begin{array}{l} {{A=3} \\ {B=-7}} \end{array}\right. \\\\\\\boxed{\dfrac{3x+2}{(x+3)^2}=\dfrac{3}{x+3}+\dfrac{-7}{(x+3)^2}}[/tex]