Inverse z transform - contour integration
In order to be able to do this, you need to know Cauchy's residue theorem, from which you get

$$\frac{1}{2\pi j}\oint_Cf(z)dz=\sum_kR_k\tag{1}$$

where $R_k$ are the residues of the analytic function $f(z)$ at its poles $p_k$, which must lie inside the closed contour $C$. The residue at pole $p_k$ is given by
$$R_k=\lim_{z\rightarrow p_k}(z-p_k)f(z)$$

In the case of the inverse $\mathcal{Z}$-transform, we have $f(z)=X(z)z^{n-1}$. Since in your example the region of convergence is given as $|z|&gt;2/3$, we know that we need to find a causal sequence $x[n]$, so we only need to consider values $n\ge 0$. For $n&gt;0$, there is only one single pole at $z=2/3$, and the corresponding residue is

$$R_0=\lim_{z\rightarrow 2/3}(z-2/3)X(z)z^{n-1}=-\frac{1}{3}\left(\frac{2}{3}\right)^{n-1}\tag{2}$$

For $n=0$ we get another pole at $z=0$ from the term $z^{n-1}$, and the corresponding residue is

$$R_1=\lim_{z\rightarrow 0}zX(z)z^{-1}=\lim_{z\rightarrow 0}X(z)=\frac12\tag{3}$$

According to (1), the value of $x[0]$ is given by the sum of the residues (2) and (3):

$$x[0]=-\frac13\left(\frac23\right)^{-1}+\frac12=0$$

For $n&gt;0$ we only need to consider the residue $R_0$, so the inverse $\mathcal{Z}$-transform is given by

$$x[n]=-\frac13\left(\frac23\right)^{n-1}u[n-1]$$

Note that this is just meant as an exercise. In practice this example is a basic transformation, which you either know or look up in a table.