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**Bruno J.** $\displaystyle y=1/3$ so $\displaystyle x=\pm \sqrt{2-y^2} = \pm \sqrt{17}/3$. Now we test the two points; we find that $\displaystyle f$ has equal value at both, with $\displaystyle f(\pm\sqrt{17}/3,1/3)=19/3$.

Another way would have been to set $\displaystyle x=\sqrt{2}\cos \theta,\ y=\sqrt{2}\sin \theta$. Then the function along the boudary is $\displaystyle F(\theta)=6\cos^2\theta+2\sqrt{2}\sin \theta$, and $\displaystyle \frac{dF}{d\theta}=-12\cos\theta\sin\theta + 2\sqrt{2}\cos\theta$ $\displaystyle =2\cos\theta(\sqrt{2}-6\sin\theta)=\sqrt{2}x(\sqrt{2}-3\sqrt{2}y)=2x(1-3y)$, so $\displaystyle F$ has extrema when $\displaystyle x=0,\ y=\pm\sqrt{2}$ or when $\displaystyle y=1/3,\ x=\pm \sqrt{17}/3$.