Let $\angle ABM=x$. Then $\angle BMC=2x$. $BC=\frac{2}{3}MC\implies MC=\frac{3}{2}BC$. Since $M$ is the midpoint of $AC$, then $AM=\frac{3}{2}BC$. So $\frac{AM}{AB}=\frac{3BC}{2AB}$. Since $\angle BMC=2x$, then $\angle AMB=180-2x$ so $\angle BAM=180-(180-2x+x)=x$. This means that $\triangle AMB$ is isosceles so $AM=MB$. Now since $BM=CM$(midpoint) then $\triangle BMC$ is also isosceles. So $\angle MCB=\angle MBC=90-x$. Notice that $\angle ABC=(90-x)+x=90$. So $\triangle ABC$ is a right triangle with hypotenuse $AC$. $AC=3BC$ so $AB=\sqrt{9BC^2-BC^2}=2BC\sqrt{2}$. Thus, \[\frac{AM}{AB}=\frac{3BC}{2AB}=\frac{3BC}{4BC\sqrt{2}}=\frac{3\sqrt{2}}{8}.\]