Given the following function:
$F(z) = \frac{-3z+2}{(z^2-z+1)}$
It seems that using the partial fractions method wont be easy because of the conjugate complex poles so, which is the proper way to solve the inverse Z transform?
I saw it could be possible to equate the whole function to the Z transform of $cos(wT)$. Is it OK?
You should have given us the ROC as well. I will assume $|z|>1$.
Rewrite it as $$\begin{align} F(z) = \frac{-3z+2}{(z^2-z+1)}&=-3\left(\frac{z}{z^2-z+1}\right)+2\left(\frac{1}{z^2-z+1}\right)\\ &=-3X(z)+2z^{-1}X(z) \end{align}$$ So after finding the inverse $z$-transform of $X(z)$, the answer will be $$f[n]=-3x[n]+2x[n-1]$$ Now to find $x[n]=\mathcal{Z}^{-1}\{X(z)=\frac{z}{z^2-z+1}=\frac{z^{-1}}{1-z^{-1}+z^{-2}}\}$ use the standard formula (no 21) in this table for $\omega_0=\frac{\pi}{3}$ to get $$x[n]=\frac{1}{\sin\left(\frac{\pi}{3}\right)}\sin\left(\frac{\pi}{3}n\right)u[n]$$