First we note that the partial derivatives vanish simultaneously at one point:
[tex]\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)[/tex]
so there is one critical point within the region [tex]D[/tex].
The Lagrangian is
[tex]L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)[/tex]
and has partial derivatives
[tex]L_x=2x+y+2\lambda x[/tex]
[tex]L_y=x+2y+2\lambda y[/tex]
[tex]L_\lambda=x^2+y^2-1[/tex]
Set each partial derivative to 0 to find the possible critical points within the disk [tex]D[/tex]. Then we notice that
[tex]yL_x=2xy+y^2+2\lambda xy=0[/tex]
[tex]xL_y=x^2+2xy+2\lambda xy=0[/tex]
[tex]L_\lambda=0\implies x^2+y^2=1[/tex]
[tex]\implies xL_y-yL_x=x^2-y^2=0[/tex]
Since [tex]x^2+y^2=1[/tex], we have
[tex]x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}[/tex]
And since [tex]x^2-y^2=0[/tex], or [tex]x^2=y^2[/tex], we also have
[tex]x=\pm\dfrac1{\sqrt2}[/tex]
So we have four possible additional critical points to consider:
[tex]f(0,0)=0[/tex]
[tex]f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32[/tex]
[tex]f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12[/tex]
[tex]f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12[/tex]
[tex]f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32[/tex]
It should be clear enough which of these correspond to the absolute extrema of [tex]f[/tex] over [tex]D[/tex].
