jmc

Expanding $(y-x)(y-2x)(y-kx),$ we get $$-2k{x}^3+(3k+2)x^{2}y-(k+3)x{y}^2+y^3.$$To make the coefficients of $x^3$ match, we multiply by $-\frac{1}{2k}.$ Then the coefficient of $x^3$ becomes 1, and the coefficient of $x^2$ becomes $$-\frac{3k+2}{2k}=-3.$$Solving for $k,$ we find $k=\boxed{\frac{2}{3}}.$