[tex]f(x,y)=x^2+xy+y^2[/tex]
[tex]\dfrac{\partial f}{\partial x}=2x+y[/tex]
[tex]\dfrac{\partial f}{\partial y}=2y+x[/tex]
Critical points occur when both partial derivatives vanish; this happens when
[tex]\begin{cases}2x+y=0\\2y+x=0\end{cases}\implies x=0,y=0[/tex]
The function has Hessian
[tex]\mathbf H(x,y)=\begin{bmatrix}\dfrac{\partial^2f}{\partial x^2}&\dfrac{\partial^2f}{\partial x\partial y}\\\\\dfrac{\partial^2f}{\partial y\partial x}&\dfrac{\partial^2f}{\partial y^2}\end{bmatrix}=\begin{bmatrix}2&1\\1&2\end{bmatrix}[/tex]
which at [tex](0,0)[/tex] has [tex]\det\mathbf H(0,0)=3>0[/tex]. And since [tex]\dfrac{\partial^2f}{\partial x^2}\bigg|_{x=0,y=0}=2>0[/tex], it follows that a local minimum occurs at [tex](0,0)[/tex] with a value of [tex]f(0,0)=0[/tex].
Meanwhile, we can parameterize the boundary by
[tex]\begin{cases}x=\cos t\\y=\sin t\end{cases}[/tex]
with [tex]0\le t\le2\pi[/tex]. So
[tex]f(x,y)=f(x(t),y(t))=F(t)=\cos^2t+\sin t\cos t+\sin^2t=1+\dfrac12\sin2t[/tex]
which has critical points when [tex]F'(t)=0[/tex]:
[tex]F'(t)=\cos2t=0\implies 2t=\dfrac\pi2+n\pi\implies t=\dfrac\pi4+\dfrac{n\pi}2[/tex]
We only have 4 points to worry about: [tex]t=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4[/tex]
Now,
[tex]F\left(\dfrac\pi4\right)=\dfrac32[/tex]
[tex]F\left(\dfrac{3\pi}4\right)=\dfrac12[/tex]
[tex]F\left(\dfrac{5\pi}4\right)=\dfrac32[/tex]
[tex]F\left(\dfrac{7\pi}4\right)=\dfrac12[/tex]
So we find that an absolute minimum occurs at [tex](0,0)[/tex] with a value of [tex]0[/tex], and two more extrema occur along the boundary when [tex]t=\dfrac\pi4[/tex] and [tex]t=\dfrac{5\pi}4[/tex], i.e at the points [tex]\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)[/tex] and [tex]\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)[/tex], both with the same maximum value of [tex]\dfrac32[/tex].
