If  $f(x+\frac{1}{2})+f(x-\frac{1}{2})=f\left(x\right)$ for all  $x\in R$, then the period of  $f\left(x\right)$ is

$f(x+\frac{1}{2})+f(x-\frac{1}{2})=f\left(x\right)\cdots \left(1\right)$

$f(x+\frac{1}{2}+\frac{1}{2})+f(x+\frac{1}{2}-\frac{1}{2})=f(x+\frac{1}{2})$

$f(x+1)+f\left(x\right)=f(x+\frac{1}{2})\cdots \left(2\right)$

$\left(1\right)+\left(2\right)$

$f(x+\frac{1}{2})+f(x-\frac{1}{2})+f(x+1)+f\left(x\right)=f\left(x\right)+f(x+\frac{1}{2})$

$f(x-\frac{1}{2})+f(x+1)=0$

Replace  $x$ as  $x+\frac{1}{2}$

$f(x+\frac{1}{2}-\frac{1}{2})+f(x+\frac{1}{2}+1)=0$

$f\left(x\right)=-f(x+\frac{3}{2})$

Replace  $x$ as  $x+\frac{3}{2}$

$f(x+\frac{3}{2})=-f(x+3)$

$\therefore f\left(x\right)=f(x+3)$

$\therefore$ Period is 3