smc

Because $64=4^3$ and $256=4^4$, we can change all of the numbers in the equation to exponents with base $4$ and solve the equation: $$\begin{array}{r}\frac{64^{x-1}}{4^{x-1}}=256^{2x}\\ \frac{(4^3{)}^{x-1}}{4^{x-1}}=(4^4{)}^{2x}\\ \frac{4^{3x-3}}{4^{x-1}}=4^{8x}\\ 4^{2x-2}=4^{8x}\\ 2x-2=8x\\ x-1=4x\\ 3x=-1\\ x=\frac{-1}{3}\end{array}$$ Thus, our desired answer is $\boxed{-\frac{1}{3}}$.