By the quadratic formula the zeroes of $z^2-z+1$ are $\frac12(1\pm i\sqrt3)$, the two non-real cube roots of $1$. Let $\omega=\frac12(1+ i\sqrt3)$; the other root is $\overline\omega$, the complex conjugate of $\omega$. Then $z^2-z+1=$ $(z-\omega)(z-\overline\omega)$.
Now just set up the usual partial fractions computation:
$\frac1{z^2-z+1}=\frac{u}{z-\omega}+\frac{v}{z-\overline\omega}\;,$ so $u(z-\overline\omega)+v(z-\omega)=1$, and you have the system $\left\{\begin{align*}&u+v=0\\&u\overline\omega+v\omega=-1\;.\end{align*}\right.$ That ought to be pretty straightforward to solve.