What would you apply?

$2013x^2 - 4008xy + 2013y^2 = 2004(x-y)^2 + 9(x^2 + y^2),$

so that, using the distance formula $D^2 = x^2 + y^2$ ,

$2004 (x-y)^2 + (3D)^2 = 1.$

The distance $D$ is maximal at the minimum $(x-y)^2$ , which obviously zero. Therefore

$3D_{max} = 1 \therefore D_{max} = \frac{1}{3}.$

it is symmetric about y=x line and the intersection with y=x will ensure that, at that point , the line y=x is normal to that curve ! And the smallest distance between the curves is along the common normal ! so putting y=x in the given curve we get a point and by using the distance formula we can find what is required !