I have a particle of mass $m$ that moves in 2-d in the potential $V(x,y)=\frac{1}{2}m\omega^2(6x^2-2xy+6y^2)$. I have to use the coordinates $u=\frac{x+y}{\sqrt 2}$ and $w=\frac{x-y}{\sqrt2}$ to show that the schrodinger equation in $(u,w)$ is given by $-\frac{\hbar^2}{2m}\left(\frac{d^2}{du^2}+\frac{d^2}{dw^2}\right)\psi(u,w)+\bar{V}(u,w)\psi(u,w) = E\psi(u,w)$
and to find $\bar{V}(u,w)$ and $\frac{d^2}{du^2},\frac{d^2}{dw^2}$.
I've done questions regarding particles moving in 2-d in the potential $V(x,y)$ before, but I'm not sure how to make the variable change as required above. I know the schrodinger equation in $(x,y)$ is given by: $-\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}\right)\psi(x,y)+V(x,y)\psi(x,y) = E\psi(x,y)$ But how would i find $\bar{V}(u,w)$ and $\frac{d^2}{du^2},\frac{d^2}{dw^2}$?
I tried rearranging $u=\frac{x+y}{\sqrt 2}$ in terms of $x$ and $y$ and substituting it into $v(x,y)$ but I still end up with $x$ and $y$ terms.