Minimum needed

Let the minimum value be $m$

$x^{2}+2xy+3y^{2}-6x-2y\geq m$

$x^{2}+2xy+3y^{2}-6x-2y-m\geq 0$

Its discriminant $D\leq 0$

$4(y-3)^{2}-4(3y^{2}-2y-m)\leq 0$

$2y^{2}+4y-(m+9)\geq 0$

$y^{2}+2y\geq \frac{m+9}{2}$

$(y+1)^{2}\geq \frac{m+11}{2}$

as least value of a perfect square is zero

so $\frac{m+11}{2}=0$

$\boxed{m=-11}$

There is a quicker way to do this with calculus, but it is still possible with only algebra:

First off, factor with respect to x.

$x^2 + 2(y - 3)x + (3y^2-2y)$

Complete the square and simplify:

$x^2 + 2(y - 3)x + (y - 3)^2 - (y - 3)^2 + (3y^2 - 2y)$

$(x + (y - 3))^2 - (y^2 - 6y + 9) + (3y^2 - 2y)$

$(x + (y - 3))^2 - y^2 + 6y - 9 + 3y^2 - 2y$

$(x + (y - 3))^2 + 2y^2 + 4y - 9$

Then, complete the square with regard to the y variables, and simplify:

$(x + (y - 3))^2 + 2y^2 + 4y - 9$

$(x + (y - 3))^2 + 2(y^2 + 2y) - 9$

$(x + (y - 3))^2 + 2(y^2 + 2y + 1 - 1) - 9$

$(x + (y - 3))^2 + 2(y^2 + 2y + 1) - 2 - 9$

$(x + (y - 3))^2 + 2(y + 1)^2 - 11$

You could stop here, since the lowest value any square number can have (in real number space) is 0, the minimum value of this equation is -11. But let's solve for the variables and make sure.

The second term contains only a $y$ variable, so if we set $y = -1$ that term is cancelled out.

This makes the first term $(x + (-1 - 3))$ or $(x - 4)$ so $x = 4$ . And at this point, the first two terms are zero, making the minimum value -11.

Checking the values in the original equation, you get:

$4^2 + 2(4)(-1) + 3(-1)^2 - 6(4) - 2(-1) = 16 - 8 + 3 - 24 + 2 = -11$

Partial derivatives will do the magic