The singularity of $\dfrac{z^2-z+1}{z-1}$ is at $z=1$. Since $|1| = 1 > \dfrac{1}{2}$, the pole $z=1$ lies outside the contour $|z|=\frac{1}{2}$. By Cauchy's theorem, the integral of an analytic function over a closed contour is $0$ when no singularities are enclosed. Therefore $\displaystyle\oint_C \frac{z^2-z+1}{z-1}\,dz = \mathbf{0}$.