Net ionic equations for an aqueous solution
Your answer is probably correct. There are some extra details one might wish to consider. If a short yes/no answer was enough, you must not read any further.

First, let us remind ourselves that in aqueous solutions metal ions coordinate with water molecules. Octahedral complexes are most common.

$$\ce{Ni^2+(aq) + 6 H2O(l) &lt;=&gt; [Ni(OH2)6]^{2+} (aq)}$$

Then $\ce{NH4OH}$, better described as a hydrate $\ce{NH3.H2O}$, is added.
$$\ce{NH3 (aq) + H2O (l)&lt;=&gt;NH3.H2O (aq)&lt;=&gt;NH4+ (aq) +OH-(aq)}$$

Now either $\ce{NH3}$ or $\ce{OH-}$ steals a proton from $\ce{[Ni(OH2)6]^{2+}}$. Note that the resulting complexes are equivalent.

$$\ce{[Ni(OH2)6]^{2+}(aq)+NH3.H2O(aq) &lt;=&gt; [Ni(OH)(OH2)5]+(aq) + NH4+(aq) +H2O(l)}$$

$$\ce{[Ni(OH2)6]^{2+}(aq) + OH-(aq)&lt;=&gt; [Ni(OH)(OH2)5]+(aq) + H2O(l)}$$

If another proton is taken, we reach the required precipitate $\ce{Ni(OH)2}$. (Due to the equivalency mentioned, I will write only one option.) 

$$\ce{[Ni(OH)(OH2)5]+ (aq) + OH-(aq)&lt;=&gt; [Ni(OH)2(OH2)4](s) + H2O (l) \tag 1}$$

Often, however, the $\ce{Ni(OH)2}$ precipitate is not visible. The whole point of this excercise was to give background to the explanation that follows. When $\ce{NH3.H2O}$ is in excess, the following processes occur. The precipitate could directly react, in which case we might end up with

$$\ce{[Ni(OH)2(OH2)4](s) + 6 NH3.H2O (aq)&lt;=&gt; [Ni(NH3)6](OH)2 (aq) + 10H2O(l)}$$

Generally, the preferred explanation is through Le Chatelier's principle. $\ce{[Ni(OH2)6]^{2+}}$ ions react via $(2)$

$$\begin{align}
\ce{[Ni(OH2)6]^{2+} + 4 NH3.H2O (aq)} &amp; \ce{&lt;=&gt; [Ni(NH3)4(OH2)2]^{2+}(aq) + 8H2O(l)}\\ 
\ce{[Ni(OH2)6]^{2+} + 6 NH3.H2O(aq) } &amp; \ce{&lt;=&gt; [Ni(NH3)6]^{2+}(aq) + 10 H2O(l)}\\
\ce{[Ni(NH3)6]^2+(aq) +2 OH-(aq) } &amp; \ce{&lt;=&gt; [Ni(NH3)6](OH)2(aq)}
\end{align}$$

Since the $\ce{[Ni(OH2)6]^{2+}}$ ions are used up, the equilibrium of $(1)$ tilts to the left, thus the precipitate disappears.

Once more, this was simply to explain why showing $\ce{Ni(OH)2}$ as a precipitate might not correspond with reality when there is (excess) dissolved ammonia.