If we rotate the $x$ and $y$ axes (along with our conic) to new axes $x'$ and $y'$, then the old coordinates, in terms of the new ones, are:

$x = x'\cos 45^{\circ} + y'\sin 45^{\circ} = \dfrac{\sqrt{2}}{2}(x' + y')$

$y = y'\cos 45^{\circ} - x'\sin 45^{\circ} = \dfrac{\sqrt{2}}{2}(y' - x')$

So the equation in our new coordinates is:

$\left( \dfrac{\sqrt{2}}{2}(x' + y')\right)^2 + \left(\dfrac{\sqrt{2}}{2}(x' + y')\right)\left(\dfrac{\sqrt{2}}{2}(y' - x')\right) + \left(\dfrac{\sqrt{2}}{2}(y' - x')\right)^2 = 3$ or:

$\dfrac{1}{2}(x'^2 + 2x'y' + y'^2) + \dfrac{1}{2}(y'^2 - x'^2) + \dfrac{1}{2}(y'^2 - 2x'y' + x'^2) = 3$, and cancelling and collecting terms:

$x'^2 + 3y'^2 = 6$

Compare these two plots:

x^2+xy+y^2=3 - Wolfram|Alpha

x^2 + 3y^2 = 6 - Wolfram|Alpha