Looks like [tex]f(x,y)=x^2+y^2-x^2y+7[/tex].
[tex]f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1[/tex]
[tex]f_y=2y-x^2=0\implies2y=x^2[/tex]
The latter two critical points occur outside of [tex]D[/tex] since [tex]|\pm\sqrt2|>1[/tex] so we ignore those points.
The Hessian matrix for this function is
[tex]H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}[/tex]
The value of its determinant at (0, 0) is [tex]\det H(0,0)=4>0[/tex], which means a minimum occurs at the point, and we have [tex]f(0,0)=7[/tex].
Now consider each boundary:
[tex]f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4[/tex]
which has 3 extreme values over the interval [tex]-1\le y\le1[/tex] of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).
[tex]f(-1,y)=8-y+y^2[/tex]
and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).
[tex]f(x,1)=8[/tex]
which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).
[tex]f(x,-1)=2x^2+8[/tex]
which has 3 extreme values, but the previous cases already include them.
Hence [tex]f(x,y)[/tex] has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).
