Z Transform How to Handle the Inital Condition

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)."

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

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<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.

Since

\$|z| = 0 ~~\Leftrightarrow~~ z = 0 \$

for z = 0 you cannot compute the term outside the sum symbol, i.e. \$z^{-1}\$, which is part of your transform. In other words, for \$z = 0\$ the transform does not exist.

Maybe you were put off-tracks because \$|z|\ge 0\$ by definition, but now you have to exclude \$ |z| = 0 \$ from the interval of valid values of \$|z|\$.