=>3x^2y-x+xy=6
=>3x^2y+xy-x=6
=>xy(3x+1)-x=6
=>xy(3x+1)-x-1/3=17/3
=>(x+1/3)(3xy-1)=17/3
=>(3x+1)(3xy-1)=17
mà x,y nguyên
nên \(\left(3x+1;3xy-1\right)\in\left\{\left(1;17\right);\left(-17;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-6;0\right)\right\}\)