jmc

Since $-\frac{5}{8}$ is a negative number, $f(x)$ is only defined for integer values of $x$, and will alternate between positive and negative values. Additionally, $\left|-\frac{5}{8}\right|<1$, so $|f(x)|$ will continually decrease and approach 0 as $x$ increases in the interval $x\ge 0$. Therefore, the largest positive value will occur at $x=0$, giving us the positive upper bound of $\lfloor {\left(-\frac{5}{8}\right)}^0\rfloor =1$. The negative value that is greatest in magnitude then occurs at the next integer value of $x$: $x=1$, giving us the negative lower bound of $\lfloor {\left(-\frac{5}{8}\right)}^1\rfloor =-1$. This tells us that $-1\le f(x)\le 1$. Since the $f(x)$ must be an integer, the only possible distinct values contained in the range are -1, 0, and 1. This gives us a total of $\boxed{3}$ values of $f(x)$ when $x\ge 0$.