Suppose that $\left|x+y\right|+\left|x-y\right|=2$. What is the maximum possible value of $x^2-6x+y^2$?
First, we find the graph of $|x + y| + |x - y| = 2$ in the coordinate plane.  To find this graph, we first consider the case where $x \ge 0$ and $y \ge 0.$  If $y \ge x,$ then
\[|x + y| + |x - y| = x + y + y - x = 2,\]so $y = 1.$

If $y \le x,$ then
\[|x + y| + |x - y| = x + y + x - y = 2,\]so $x = 1.$  Thus, the graph of in the first quadrant consists of the line segment connecting $(1,0)$ to $(1,1),$ and the line segment connecting $(0,1)$ to $(1,1).$

Now, let $(a,b)$ be a point on the graph, so
\[|a + b| + |a - b| = 2.\]Then for $x = a$ and $y = -b,$
\[|x + y| + |x - y| = |a - b| + |a + b| = 2.\]This shows that if $(a,b)$ is a point on the graph, then so is $(a,-b).$  Thus, the graph is symmetric over the $x$-axis.

Similarly, we can show that if $(a,b)$ is a point on the graph, then so is $(-a,b)$.  Thus, the graph is also symmetric over the $y$-axis.  Hence, the graph is the square with vertices $(1,1),$ $(-1,1),$ $(-1,-1),$ and $(1,-1).$

[asy]
unitsize(2 cm);

pair A, B, C, D;

A = (1,1);
B = (-1,1);
C = (-1,-1);
D = (1,-1);

draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));
draw(A--B--C--D--cycle);

label("$(1,1)$", A, NE);
label("$(-1,1)$", B, NW);
label("$(-1,-1)$", C, SW);
label("$(1,-1)$", D, SE);
[/asy]

Notice that $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$ is equivalent to 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 point, when substituting into the function, yields $\boxed{8}$.