\(A=\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{9x^2-y^2}\\ A=\dfrac{9x^2-3xy-3x^2-xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}\\ x=1;y=2\Leftrightarrow A=\dfrac{6-8+2}{\left(3-2\right)\left(3+2\right)}=\dfrac{0}{5}=0\)