Constrained extreme problem

I have some trouble with this exercise. It says as follows: Find the constrained extremes for $\displaystyle z=2x^2+y^2$ over $\displaystyle \cos (x^2+y^2)-1=0$

I think that difficult of this problem is that I have a family of circles that satisfies $\displaystyle \cos (x^2+y^2)-1=0$, the family of circles with radius $\displaystyle r=\sqrt[ ]{k \pi}$.

This is what I did.

$\displaystyle \cos (x^2+y^2)-1=0\Rightarrow{x^2+y^2=k \pi},k\in{Z}$

$\displaystyle F(x,y,\lambda)=2x^2+y^2-\lambda(cos(x^2+y^2)-1)=0$

$\displaystyle \begin{Bmatrix}F_x=4x+2\lambda x \sin(x^2+y^2)=0 & & (1)\\F_y=2y+2\lambda y \sin(x^2+y^2)=0 & & (2)\\F_{\lambda}=\cos(x^2+y^2)-1=0 & & (3)\end{matrix}$

From (1) and (2): $\displaystyle 4x-2y+(2\lambda \sin(x^2+y^2))(x-y)=0\Longrightarrow{2\lambda\sin(x^2+y^2)=\display style\frac{4x-2y}{x-y}}$ (*)

Replacing with (*) in (1)

$\displaystyle 4x+x \displaystyle\frac{4x-2y}{x-y}}=0\Rightarrow{4x(x-y)+x(4x-2y)=0}\Longrightarrow{x^2-4xy=0\longrightarrow{y=\displaystyle\frac{-x}{4}}}$

Replacing in (3)

$\displaystyle x^2+(\displaystyle\frac{-x}{4})^2=k\pi\Rightarrow{x=\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}$

Well, I'm not sure if this is okey. If it is, how should I proceed?

Thanks for posting!