The given equation is - $$3x^2 + 2xy + 3y^2 - 32y +92=0$$
To get rid of xy term i used the substitutions - $$x=p+q , y=q-p$$
Then the equation becomes - $$(p-4)^2 + 2(q-2)^2=1$$ which is an ellipse with center at $p=4$ and $q=2$.
So in the original system , $x=p+q=6$ and $y=q-p=-2$ .
But the actual center as wolfram alpha shows is the opposite , that is $x=-2$ and $y=6$.
I don't know why this is happening and i'm really confused.
Please help.
Hint
I did what you tell $(x=p+q , y=q-p)$ and the raw result I obtained for the expression is $$4 p^2+32 p+8 q^2-32 q+92=0$$ I let you the task of grouping but it seems that you made some mistakes.
I am sure that you can take from here and fix the problem.