1. We are given f(x+y)=f(x)f(y)
   for all }x,y\in\mathbb R, with $f(x)\neq 0$ for every $x\in\mathbb R$.

2. Also, $f$ is differentiable at $x=0$ and $f'(0)=3.$

3. We need to find $\lim_{h\to 0}\frac{f(h)-1}{h}.$

4. But by the definition of derivative at $x=0$, $f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}.$ So we first need $f(0)$.

5. Put $x=0$ and $y=0$ in the functional equation: $f(0)=f(0+0)=f(0)f(0)=f(0)^2.$ Hence, $f(0)^2-f(0)=0$ $f(0)(f(0)-1)=0.$ Since $f(0)\neq 0$, we must have $f(0)=1.$

6. Therefore, $f'(0)=\lim_{h\to 0}\frac{f(h)-1}{h}.$

7. Given that $f'(0)=3$, it follows directly that $\lim_{h\to 0}\frac{f(h)-1}{h}=3.$

Therefore, the required integer is $\boxed{3}.$