jmc

Let $y$ be a number in the range of $f.$ This means that there is a real number $x$ such that $$y=\frac{x}{x^2-x+1}.$$Multiplying both sides by $x^2-x+1$ and rearranging, we get the equation $$y{x}^2-(y+1)x+y=0.$$Since $x^2-x+1=(x-\frac{1}{2}{)}^2+\frac{3}{4}>0$ for all $x,$ our steps are reversible, so $y$ is in the range of $f$ if and only if this equation has a real solution for $x.$ In turn, this equation has a real solution for $x$ if and only if the discriminant of this quadratic is nonnegative. Therefore, the range of $f$ consists exactly of the values of $y$ which satisfy $$(y+1{)}^2-4y^2\ge 0,$$or $$0\ge 3y^2-2y-1.$$This quadratic factors as $$0\ge (3y+1)(y-1),$$which means that the solutions to the inequality are given by $-\frac{1}{3}\le y\le 1.$ Therefore, the range of $g$ is the closed interval $\boxed{[-\frac{1}{3},1]}.$