$3y^2+x^2+2xy+2x+6y+3=0$
⇔ $y^2+2y^2+x^2+2xy+2x+2y+4y+1+2=0$
⇔ $(x^2+y^2+1+2x+2y+2xy)+(2y^2+4y+2)=0$
⇔ $(x+y+1)^2+2(y+1)^2=0$
$Vì \left \{ {{(x+y+1)^2\geq0 } \atop {2(y+1)^2\geq0}} \right.,$ $với$ $mọi$ $x,y$ $nên$ $(x+y+1)^2+2(y+1)^2=0$
⇔ $\left \{ {{(x+y+1)^2=0 } \atop {2(y+1)^2=0}} \right.,$
⇔ $\left \{ {{x+y+1=0 } \atop {y+1=0}} \right.,$
⇔ $\left \{ {{x+y=-1 } \atop {y=-1}} \right.,$
⇔ $\left \{ {{x=0 } \atop {y=-1}} \right.,$
$Vậy$ $x=0, y=-1$
GOOD LUCK!
