jmc

We try graphing both equations in the $xy-$plane. The graph of $x+3y=3$ is a line passing through $(3,0)$ and $(0,1).$ To graph $\left||x|-|y|\right|=1,$ notice that the equation does not change if we replace $x$ by $-x$ or if we replace $y$ by $-y.$ Thus, the graph of $\left||x|-|y|\right|=1$ is symmetric about the $y-$axis and the $x-$axis, so if we graph the equation in just the first quadrant, we can produce the remainder of the graph by reflecting it over the axes.

If $(x,y)$ lies in the first quadrant, then $x\ge 0$ and $y\ge 0,$ so the equation $\left||x|-|y|\right|=1$ becomes just $|x-y|=1.$ Therefore, either $x-y=1$ or $y-x=1,$ whose graphs in the first quadrant are rays. This gives us the whole graph of $\left||x|-|y|\right|=1:$ (The graph of $\left||x|-|y|\right|=1$ is drawn in blue, and the line $x+3y=3$ is drawn in red.) We see that the two graphs intersect at $\boxed{3}$ points.