This is what I have to far...
Let $f(x)=g \implies gcd(x,k)=g$
We can then write, $ax+bk=g$
Let $f(y)=h \implies gcd(y,k)=h$
We can write, $cy+dk=h$
Multiplying, $(ax+bk)(cy+dk)=gh$
$\implies axcy+axdk+cybk+bdkk=gh$
$\implies acxy+k(adx+cby+bdk)=gh$
Let $r=ac$ and $s=adx+cby+bdk$
$\implies rxy+sk=gh$
Thus, $gcd(xy,k)=gh$
$\implies f(xy)=f(x) \cdot f(y)$
Hence, $f(n)=gcd(n,k)$ is multiplicative.
I was just wondering if this looked good...
Note that for $x,y \in \mathbb{Z}$ if we have $ax + by = c$ for $a,b \in \mathbb{Z}$ we cannot conclude that $\gcd(x,y) = c$, rather we can say that $\gcd(x,y) | c$. Thus in your proof we have $\gcd(xy,k) | gh$. Clearly we have $g | \gcd(xy,k)$ and $h | \gcd(xy,k)$. Let $\gcd(g,h) = l$. Then $l | g$ and $l | h$. Since $g | x$ and $h | y$ we have $l | x$ and $l | y$ so $l | \gcd(x,y) = 1$ thus $l = 1$. Therefore $gh | \gcd(xy,k)$ and since $\gcd(xy,k) | gh$ we can now conclude that $gh = \gcd(xy,k)$ so $f(xy) = f(x)f(y)$.