Inverze z transform - contour integration
From definition of inverse z-transforn,


  "The inverse Z-transform is,
  $$x(n) = \frac{1}{2\pi j}\oint_{C}^{ } X(z)z^{n-1}dz$$
  where C is a counterclockwise closed path encircling the origin and entirely in the region of convergence (ROC). In the case where the ROC is causal this means the path C must encircle all of the poles of X(z)."


$$\begin{align} x(n)&amp;= -\frac{1}{3}\times \frac{1}{2\pi j}\oint_{C}^{ } \frac{z^{n-1}}{z-\frac{2}{3}}dz\\
&amp;= -\frac{1}{3}\times \frac{1}{2\pi j}\oint_{C}^{ } \frac{f(z)}{z-\frac{2}{3}}dz\end{align}$$

Where, \$f(z) = z^{n-1}\$. Using Cauchy's Integral formula:

$$x(n) = -\frac{1}{3}\times \left(\dfrac{2}{3}\right)^{n-1}$$

If \$n&lt;1\$, then the function \$f(z) = z^{n-1}\$ will have singularity at \$z=0\$ and hence Cauchy's Integral formula can not be applied. So \$n\$ must be greater than or equal to \$1\$. Or we can write:

$$x(n) = -\frac{1}{3}\times \left(\dfrac{2}{3}\right)^{n-1}u(n-1)$$

PS: Read this page. They have given equations to find inverse z-transform directly (without using integration) using residue method.