Answer
$f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=2$ and $y=-2.$
Work Step by Step
First we will find both partial derivatives. The partial derivative with respect to $x$ is:
$$f_x(x,y)=\frac{\partial}{\partial x}(x^2+xy+y^2-2x+2y)=2x+y-2$$
The partial derivative with respect to $y$ is:
$$f_y(x,y)=\frac{\partial}{\partial y}(x^2+xy+y^2-2x+2y)=x+2y+2$$
To find all the values of $x$ and $y$ such that $f_x(x,y)=0$ and $f_y(x,y)=0$ we will equate both partial derivatives with 0:
$$f_x(x,y)=2x+y-2=0,f_y(x,y)=x+2y+2=0$$
From the second equation we have that $x=-2y-2$, so putting this into the first equation we get:
$$2(-2y-2)+y-2=0\Rightarrow-3y-6=0\Rightarrow y=-2$$
Now for $x$ we have:
$$x=-2y-2=-2\cdot(-2)-2=2$$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=2$ and $y=-2.$
