#### Answer

The critical points are $y=0, 2$.

#### Work Step by Step

Given, $g(y) = \frac{y-1}{y^2-y+1}$. Differentiating $g$ with respect to $y$ using quotient's rule, $$\frac{dg}{dy}=\frac{(1)(y^2-y+1)-(y-1)(2y-1)}{(y^2-y+1)^2}$$
The term $\frac{dg}{dy}$ is equal to $0$ at critical numbers. Thus, $$\frac{(1)(y^2-y+1)-(y-1)(2y-1)}{(y^2-y+1)^2}=0$$ $$y^2-y+1-2y^2+y+2y-1=0$$ $$-y^2+2y=0$$ $$y=0, 2.$$