1. Given $f(x)=x^{2}-6 x+5$, find the range of points $(x, y)$ on the plane that satisfy $f(x)+f(y) \leqslant 0$ and $f(x)-f(y) \geqslant 0$. Also, draw the graph.
$f(x)+f(y)=x^{2}+y^{2}-6 x-6 y+10 \leqslant 0 \Rightarrow (x-3)^{2}+(y-3)^{2} \leqslant 8$, then the points $(x, y)$ satisfying $f(x)+f(y) \leqslant 0$ are located within the circle (including the boundary) in the plane with center $(3,3)$ and radius $2 \sqrt{2}$;
$$
f(x)-f(y)=x^{2}-y^{2}-6(x-y) \geqslant 0
$$
$\Rightarrow (x-y)(x+y-6) \geqslant 0$, then the points $(x, y)$ satisfying
$f(x)-f(y) \geqslant 0$ are located in the regions (including the boundaries) on either side of the two intersecting lines $x-y=0$ and $x+y-6=0$.
In summary, the points $(x, y)$ satisfying the conditions are located in the shaded area shown in the figure.