jmc

The expansion of $(x+m{)}^2+\frac{1}{12}$ is $x^2+2mx+m^2+\frac{1}{12}$, which has a constant term of $m^2+\frac{1}{12}$. This constant term must be equal to the constant term of the original quadratic, so $$m^2+\frac{1}{12}=\frac{1}{3},$$and $$m^2=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}.$$This yields the possibilities $m=\frac{1}{2}$ and $m=-\frac{1}{2}$.

If $m=\frac{1}{2}$, then $(x+m{)}^2+\frac{1}{12}=x^2+x+\frac{1}{4}+\frac{1}{12}=x^2+x+\frac{1}{3}$. This implies $b=1$, but we reject this possibility because we were told that $b$ is a negative number.

If $m=-\frac{1}{2}$, then $(x+m{)}^2+\frac{1}{12}=x^2-x+\frac{1}{4}+\frac{1}{12}=x^2-x+\frac{1}{3}$, giving the result $b=\boxed{-1}$.