smc

Plugging in some values, we see that the graph of the equation $|x+y|+|x-y|=2$ is a square bounded by $x=\pm 1$ and $y=\pm 1$. Notice that $x^2-6x+y^2=(x-3{)}^2+y^2-9$ means the square of the distance from a point $(x,y)$ to point $(3,0)$ minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point $(3,0)$, which is $(-1,\pm 1)$. Either one, when substituting into the function, yields $\boxed{8}$.