Answer
$f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=3$ and $y=1.$
Work Step by Step
First we will find both partial derivatives. The partial derivative with respect to $x$ is:
$$f_x(x,y)=\frac{\partial }{\partial x}(x^2-xy+y^2-5x+y)=2x-y-5$$
The partial derivative with respect to $y$ is:
$$f_y(x,y)=\frac{\partial }{\partial y}(x^2-xy+y^2-5x+y)=-x+2y+1$$
To find all 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 $0$:
$$f_x(x,y)=2x-y-5=0,f_y(x,y)=-x+2y+1=0$$
From the second equation we have that $x=2y+1$. Putting this into the first equation we get:
$$2(2y+1)-y-5=0\Rightarrow3y-3=0\Rightarrow y=1$$
Now we for $x$ have:
$$x=2y+1=2\cdot1+1=3$$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=3$ and $y=1.$
