$(1)$ $\mathbb{Z}_{n}$ is cyclic and has a subgroup of order $m$ only if $m$ divides $n$ (by the theorem of Lagrange).
$(2)$ Any subgroup of a cyclic group is cyclic.
$(3)$ In your problem, any subgroup of the form $D\le G = \{0, d\}$ $d\ne 0$, $d \in \mathbb{Z}_n$ clearly has order $2$ and must be cyclic, so $D = \langle d \rangle$.
So yes, $d+d = 0$, and hence $d^{-1} = -d = d$, otherwise $|\langle d\rangle| = \{0, d\}\ne 2$.
By $(1)$, $|\langle d \rangle| = 2$ must divide $n.\;$ So $\;n = 2k\;$ where k is some positive integer. Hence $n$ must be even.
Now, what can you say about the possible element $d$ such that $d$ generates a subgroup of order $2$, given that $|\mathbb{Z}_{n}|$ is even $n = 2k$?
$|G| = 2:\;\;$ If $G = \mathbb{Z}_2$, $D = \{0, 1\} = G$.
$|G| > 2: \;\;k = d \;\; \implies\quad n = 2k = 2d \implies d = \dfrac{n}{2}.$
Try specific examples to get clear on what's going on here: What element in $\mathbb{Z}_4$, generates a subgroup of order $2$? Likewise, for $Z_{12}$ what is the element that generates a subgroup of order $2$?