1. We are given

   \quad \text{for all } x,y\in \mathbb N,$$ with $$f(1)=3.$$

2. Use the functional equation to find $f(n)$ for natural numbers.

   Put $y=1$: $f(x+1)=f(x)f(1)=3f(x).$ So the function forms a geometric progression on natural numbers.

   In particular, $f(2)=f(1)f(1)=3\cdot 3=9=3^2,$ $f(3)=f(2)f(1)=9\cdot 3=27=3^3.$

   By induction, $f(n)=3^n \quad \text{for all } n\in \mathbb N.$

3. Now use the given sum: $\sum_{k=1}^{n} f(k)=3279.$ Since $f(k)=3^k$, we get $\sum_{k=1}^{n} 3^k=3279.$

4. This is a geometric series: $\sum_{k=1}^{n} 3^k=\frac{3(3^n-1)}{3-1}=\frac{3^{n+1}-3}{2}.$

   Hence, $\frac{3^{n+1}-3}{2}=3279.$

5. Solve for $n$: $3^{n+1}-3=6558,$ $3^{n+1}=6561.$

   Now, $6561=3^8.$ Therefore, $n+1=8 \implies n=7.$

6. Check with options:

   • A: $9$
   • B: $7$ ✅
   • C: $6$
   • D: $8$

So the correct option is B.