Let 
$$
L=\mathbb{Q}(\sqrt{-1})\otimes\_\mathbb{Q} \mathbb{Q}\_3
$$
where $\mathbb{Q}\_3$ denotes de $3$-adic rational numbers. 
Then $L$ is a quadratic extension of the local field $\mathbb{Q}\_3$. 
Furthermore, the valuation ring of $L$ is $B:=\mathbb{Z}[\sqrt{-1}] \otimes \mathbb{Z}\_3$.


It is known that the norm map $N$ maps the set $U\_L$ of units in $B$ onto the set $U\_{\mathbb{Q}\_3}$ of units in $\mathbb{Z}\_3$ (see Serre's book "Local Fields", Chapter V, Prop. 3). 
This implies that there is an element $x\in U\_L$ such that $N(x)=-1$ since $-1$ is a unit in $\mathbb{Q}\_3$. 
Could someone specify this element $x$?



Norm map and units in local rings
One approach is by Hensel's Lemma: consider $N(x+i)=x^2+1=-1$, for example. Since $1^2+1^2=-1 \mod 3$ while $2\cdot 1=2\not= 0 \mod 3$, the equation $x^2+1=-1$ has a solution in $\mathbb Z\_3$. Hensel's lemma gives a sequence of integers approaching the solution. Note that there are many things in $\mathbb Q\_3(i)$ with norm $1$, so there's no unique solution to $N(x+iy)=-1$.


Edit: As @Lubin notes, Hensel (and also exponential/logarithm) considerations demonstrate a $\sqrt{-2}$ in $\mathbb Q\_3$, call it $\beta$. Then $N(\beta+i)=\beta^2+1=-2+1=-1$, yet again.


A point neglected in my earlier answer was that the $3$-adic exponential and log can give other ways to express things such as $\sqrt{-2}$, as alternative to Hensel. The outcomes have different utilities.


