smc

The first condition gives us that $$x^2+6x+1+y^2+6y+1\le 0⟹(x+3{)}^2+(y+3{)}^2\le 16$$ which is a circle centered at $(-3,-3)$ with radius $4$. The second condition gives us that $$x^2+6x+1-y^2-6y-1\le 0⟹(x^2-y^2)+6(x-y)\le 0⟹(x-y)(x+y+6)\le 0$$ Thus either $$x-y\ge 0,x+y+6\le 0$$ or $$x-y\le 0,x+y+6\ge 0$$ Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$, as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2}(\pi \cdot 4^2)=8\pi \approx \boxed{25}$.