Given that $f(x^2+x+1)=f(x^2-x+1)$ for all $x$, is $f(x)$ periodic? 
Given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying
  $$f(x^2+x+1)=f(x^2-x+1)\;\;\; \forall\;\;\;x\in\mathbb{R}$$ Is $f(x)$
  periodic?

My Attempt:
$$f((x+\frac{1}{2})^2+\frac{3}{4})=f((x-\frac{1}{2})^2+\frac{3}{4})$$
Let $x-\frac{1}{2}=t$
$$f((t+1)^2+\frac{3}{4})=f(t^2+\frac{3}{4})$$
Let $g(t)$ be a polynomial defined as
$$g(t)=t^2+\frac{3}{4}$$
Can I conclude anything from this?

$$f((t+1)^2+\frac{3}{4})=f(t^2+\frac{3}{4})$$
Definition of  a  periodic function $f(x+a)=f(x)$
As you can see in your equation $a$ is 1
or you can do it like this $$f(x^2+x+1)=f(x^2-x+1)\;\;\; \forall\;\;\;x\in\mathbb{R}$$
Now put $x+1$ in the second equation in place of $x$  they both become equal which means 
$$g(x+1)=g(x)$$ Thus  period of your function $g(x)=f(x^2-x+1) $ is 1
