Defining $\vec{r}=xi+yj+zk\ $ and $|\vec{r}|=r$, if $f$ is a scalar function such that $\nabla^2 f(r)=0$, then prove that $$f(r)=a+\dfrac{b}{r}$$where $a$ and $b$ are constants.
I found that $$\nabla^2 f(r)=f''(r)+\dfrac{2}{r}f'(r)=0$$
What should I do further please help?
HINT:
Note that if $f$ is a function if $r$ only, then
$$\nabla^2 f(r)=\frac1{r}\frac{\partial^2}{\partial r^2}(r f(r)) \tag1$$
Set $(1)$ equal to $0$ and integrate twice. Can you finish?