jmc

Suppose that $y$ is a fixed number, and $x$ can vary. If we try to complete the square in $x,$ we would write $$x^2+(2y-6)x+\cdots ,$$so the square would be of the form $(x+(y-3){)}^2.$ Hence, for a fixed value of $y,$ the expression is minimized in $x$ for $x=3-y.$

Setting $x=3-y,$ we get $$\begin{array}{rl}{x}^2+2xy+3y^2-6x-2y& =(3-y{)}^2+2(3-y)y+3y^2-6(3-y)-2y\\ & =2y^2+4y-9\\ & =2(y+1{)}^2-11.\end{array}$$Hence, the minimum value is $\boxed{-11},$ which occurs when $x=4$ and $y=-1.$