Given functional equation

$f(x+y)=f(x)f(y)\,\forall x,y$

Also given: $f(5)=2, \qquad f'(0)=3$

We need to find $f'(5)$.

Differentiate the functional equation with respect to $x$

Starting from $f(x+y)=f(x)f(y)$

Treat $y$ as constant and differentiate both sides with respect to $x$: $\frac{d}{dx}f(x+y)=\frac{d}{dx}[f(x)f(y)]$

Since $f(y)$ is constant with respect to $x$, $f'(x+y)=f'(x)f(y)$

Put $x=0$

Then $f'(y)=f'(0)f(y)$

Using $f'(0)=3$, $f'(y)=3f(y)$

Now evaluate at $y=5$

$f'(5)=3f(5)$

Given $f(5)=2$, $f'(5)=3\times 2=6$

Check with options

The correct option is: $\boxed{6}$

So, Option C is correct.