jmc

We take cases on the sign of $y-1.$ If $y\ge 1,$ then the equation simplifies to $$\sqrt{x^2+y^2}+(y-1)=3,$$or $$\sqrt{x^2+y^2}=4-y.$$Squaring both sides, we get $x^2+y^2=(4-y{)}^2=y^2-8y+16,$ or $x^2=-8y+16.$ Solving for $y,$ we get $$y=-\frac{1}{8}{x}^2+2,$$so the vertex of this parabola is $(0,2).$

If $y<1,$ then we have $$\sqrt{x^2+y^2}+(1-y)=3,$$or $$\sqrt{x^2+y^2}=y+2.$$Squaring both sides, we get $x^2+y^2=y^2+4y+4,$ and solving for $y$ gives $$y=\frac{1}{4}{x}^2-1.$$So the vertex of this parabola is $(0,-1).$

Thus, the distance between the vertices of the two parabolas is $|2-(-1)|=\boxed{3}.$ (Note that $3$ is the number that appears on the right-hand side. Is this a coincidence?)