We determine all the stationary points of $z(x,y)=5x-x^2-y^2+xy$.
- $z'_x(x,y)=5-2x+y$,
- $z'_y(x,y)=-2y+x$.
$z'_y(x,y)=0$ gives $x=2y$. We plug this into $z'(x,y)=0$, which gives $5-2(2y)+y=5-3y=0$. Hence, $y=\frac{5}{3}$ and $x=2y=2\frac{5}{3}=\frac{10}{3}$.
Hence, the stationary point is $(x,y)=(\frac{10}{3},\frac{5}{3})$.