HINT:

Write $H(z)$ as

$$H(z)=G(z^2)\tag{1}$$

with $$G(z)=\frac{z^{-1}}{1-0.5z^{-1}}\tag{2}$$

Determine the inverse transform $g[n]$ of $G(z)$. Figure out what replacing $z$ by $z^2$ means in the time domain. From this, the inverse transform of $H(z)$ is easily obtained from $g[n]$.

HINT:

Write $H(z)$ as

$$H(z)=G(z^2)\tag{1}$$

with $$G(z)=\frac{z^{-1}}{1-0.5z^{-1}}\tag{2}$$

Determine the inverse transform $g[n]$ of $G(z)$. Figure out what replacing $z$ by $z^2$ means in the time domain. From this, the inverse transform of $H(z)$ is easily obtained from $g[n]$.

EDIT: Your partial fraction expansion lacks the factor $z^{-2}$, and the major mistake is that you transform both terms to the same sequence, even though the poles have opposite signs. If you do it right, the two terms of the sequence only cancel each other for odd $n$, and you can combine them for even $n$. The method shown as a hint in this answer is probably more straightforward because it avoids partial fraction expansion.