jmc

Note that $$f(x)+f(y)=x^2+6x+y^2+6y+2=(x+3{)}^2+(y+3{)}^2-16$$and $$f(x)-f(y)=x^2-y^2+6(x-y)=(x-y)(x+y+6).$$The given conditions can be written as $$(x+3{)}^2+(y+3{)}^2\le 16\text{and}(x-y)(x+y+6)\le 0.$$The first inequality describes the region on and inside the circle of radius 4 with center $(-3,-3)$. The second inequality can be rewritten as $$(x-y\ge 0\text{ and }x+y+6\le 0)\text{or}(x-y\le 0\text{ and }x+y+6\ge 0).$$Each of these inequalities describes a half-plane bounded by a line that passes through $(-3,-3)$ and has slope 1 or $-1$. Thus, the set $R$ is the shaded region in the following diagram, and its area is half the area of the circle, which is $8\pi \approx \mathrm{25.13.}$, so the closest integer to $R$ is $\boxed{25}$.