Given functional equation

Find $f(0)$

Put $x=0, y=0$ in the functional equation:

$5f(0)=f(0)^2$

So,

$f(0)(f(0)-5)=0$

Since $f(0)>0$, we must have

$f(0)=5$

Reduce to an exponential form

Define

$g(x)=\frac{f(x)}{5}$

Then the given equation becomes

$5f(x+y)=f(x)f(y)$ $5\cdot 5g(x+y)=5g(x)\cdot 5g(y)$ $25g(x+y)=25g(x)g(y)$

Hence,

$g(x+y)=g(x)g(y)$

Also, since $f$ is differentiable, $g$ is differentiable. A differentiable solution of

$g(x+y)=g(x)g(y), \quad g(x)>0$

must be of the form

$g(x)=a^x$

for some $a>0$.

Therefore,

$f(x)=5a^x$

Use the value $f(3)=320$

$f(3)=5a^3=320$

So,

$a^3=64$ $a=4$

Thus,

$f(x)=5\cdot 4^x$

Compute the required sum

We need

$\sum_{n=0}^{5} f(n)=\sum_{n=0}^{5} 5\cdot 4^n=5\sum_{n=0}^{5}4^n$

Now,

$\sum_{n=0}^{5}4^n=1+4+16+64+256+1024=1365$

Therefore,

$\sum_{n=0}^{5} f(n)=5\cdot 1365=6825$

Check options

The correct option is:

$\boxed{\text{D: }6825}$