$f(x+1/2)+f(x-1/2)= f(x)$ Then the period of $f(x)$ is?

We may conclude that $f(x)=f(x+3)$ holds for all $x$.

Following @Mike Earnest's suggestion, \begin{align} f(x)&=f(x+1/2)+f(x-1/2),\\ f(x-1/2)&=f(x)+f(x-1). \end{align} Add up these two equations, and you will have $$ f(x+1/2)+f(x-1)=0, $$ or, due to the arbitrariness of $x$, $$ f(x+3/2)+f(x)=0. $$ Now let $x\to x+3/2$, and $$ f(x+3)+f(x+3/2)=0. $$ The difference of the last two equations gives $$ f(x)=f(x+3). $$

Of course, as @Przemysław Scherwentke has mentioned, this might not mean that $3$ is the period of $f$, as we do not know if this $3$ is the smallest non-negative value $T$ that ensures $f(x)=f(x+T)$. All in all, this is what we can get from the given relation.

Let $L$ be the operator on functions on $\mathbb{R}$ such that $(Lg)(x) = g(x+1/2)$. In terms of $L$, the equation at hand $f(x+1/2) + f(x-1/2) = f(x)$ is equivalent to $(L^2-L+1)f = 0$.

Notice the factorization of polynomial $$(t^6-1) = (t^3-1)(t^3+1) = (t^3-1)(t+1)(t^2-t+1)$$ Any solution of $(L^2-L+1)f = 0$ automatically satisfy $(L^6-1)f = 0$ or equivalently $f(x+3) = f(x)$. This means $3$ is a period (through not necessary the smallest period) of $f$.

For non-trivial examples, take $f(x) = \sin\left(\frac{2\pi}{3}(6k+1 )x\right)$ where $k \in \mathbb{Z}$, the smallest period of $f(x)$ is $\frac{3}{|6k+1|}$ and you can verify they are solutions to the equation at hand.

$f(x+1) + f(x) = f(x + \frac 12)= f(x) - f(x - \frac 12)$

$f(x+1) = -f(x - \frac 12)$

So $f(x - \frac 12) = -f(x - \frac 12 - 1\frac 12) = -f(x-2)$

So $f(x+1) = -f(x-\frac 12) = f(x-2)$.

So period is at most $3$.

Not sure off the top of my head how to show it doesn't necessarily need to be smaller.