[tex]f(x,y)=x^2+xy+y^2+y[/tex]
[tex]\nabla f(x,y)=(f_x,f_y)=(2x+y,x+2y+1)[/tex]
Set both partial derivatives equal to zero and solve for [tex]x,y[/tex].
[tex]\begin{cases}2x+y=0\\x+2y+1=0\end{cases}\implies x=\dfrac13,y=-\dfrac23[/tex]
The Hessian matrix (matrix of second-order partial derivatives) for this function is
[tex]H=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\implies\det H=3>0[/tex]
Now since [tex]\det H>0[/tex] for any [tex](x,y)[/tex], and [tex]f_xx=2>0[/tex] for any [tex](x,y)[/tex], it follows by the second partial derivative test that the critical point [tex]\left(\dfrac13,-\dfrac23\right)[/tex] is the site of a local minimum of [tex]f(x,y)[/tex]. The value of the minimum is [tex]f\left(\dfrac13,-\dfrac23\right)=-\dfrac13[/tex].
