jmc

We recognize that $64x^6-729y^6=(4x^2{)}^3-(9y^2{)}^3$, allowing us to first apply the difference of squares factorization, followed by the sum and difference of cubes factorizations: $$\begin{array}{rl}64x^6-729y^6& =(8x^3-27y^3)(8x^3+27y^3)\\ & =(2x-3y)(4x^2+6xy+9y^2)(2x+3y)(4x^2-6xy+9y^2)\end{array}$$The sum of all the coefficients is $2+(-3)+4+6+9+2+3+4+(-6)+9=\boxed{30}$.