Causal Inverse Z-Transform of Fibonacci

Using partial fraction techniques and denoting the zeroes of $z^2-z-1$ by $\alpha_1$ and $\alpha_2$, we have

$$ Y(z) = \frac{z}{z^2-z-1} \implies \frac{Y(z)}{z} = \left(-\frac{1}{5}+\frac{2}{5}\alpha_1\right) \frac{1}{z-\alpha_1}+ \left(-\frac{1}{5}+\frac{2}{5}\alpha_2\right) \frac{1}{z-\alpha_2} $$

$$ \implies Y(z) = \beta_1 \frac{z}{z-\alpha_1}+\beta_2 \frac{z}{z-\alpha_2} $$

$$ \implies Y(z) = \beta_1\left(1-\frac{\alpha_1}{z}\right)^{-1}+\beta_2\left(1-\frac{\alpha_2}{z}\right)^{-1} $$

where $\beta_1 =\left(-\frac{1}{5}+\frac{2}{5}\alpha_1\right)$ and $\beta_2=\left(-\frac{1}{5}+\frac{2}{5}\alpha_2\right)$.

$$ \implies Y(z) = \beta_1 \sum_{n=0}^{\infty}\frac{\alpha_1^n}{z^n}+\beta_2 \sum_{n=0}^{\infty}\frac{\alpha_2^n}{z^n}$$

$$ \implies y(n)= \beta_1\alpha_1^n + \beta_2\alpha_2^n . $$

Note: You can use the inverse Z-transform formula which uses the contour integration

$$ y(n)=\frac{1}{2\pi i}\int_{C} z^{k-1}Y(z) dz .$$