jmc

Applying $f$ to both sides of the equation $f^{-1}(x)=f((2x{)}^{-1})$, we get $f(f^{-1}(x))=f(f((2x{)}^{-1}))$. By definition of the inverse function, $f(f^{-1}(x))=x$, and $$f(f((2x{)}^{-1}))=f\left(f\left(\frac{1}{2x}\right)\right)=f\left(\frac{16}{2x}+3\right)=f\left(\frac{8}{x}+3\right)=f\left(\frac{3x+8}{x}\right)=16\cdot \frac{3x+8}{x}+3=\frac{51x+128}{x}.$$Hence, $$x=\frac{51x+128}{x}.$$Then $x^2=51x+128$, or $x^2-51x-128=0$. Vieta's formula tells us that the sum of the roots of a quadratic $a{x}^2+bx+c$ is $-\frac{b}{a}$, so in this case, the sum of the roots is $\boxed{51}$.