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Complete the square:The minimum locus of a squared function occurs where its residual vanishes. Hence the minimum is $-6$ on $y=-x^2/2$, provided these points lie in the ellipse. Setting $t=x^2$ in the constraint gives $t+t^2/2\le 9$, or $0\le t\le \sqrt{19}-1$. ThusFor the maximum, let $q=x^2+2y$. The ellipse givesAlso $q\ge 2y\ge -3\sqrt{2}$, whose absolute value is less than $19/2$. Therefore $q^2\le (19/2{)}^2$. Equality is attained at $y=1/2$, $x^2=17/2$, givingThese attained bounds are global, not merely stationary-point tests. The minimizing locus is a parabola segment.