So we would have $f(x) = f(-x)$ and $f$ can either be ascending or descending.
My approach to the problem would be by Reduction To Absurdity. I would start off by saying that: "Suppose there actually is a function $f$ that meets those 2 requirements and is not constant". To make things easier I'd probably make a choice and say we're working with an increasing function. We then take two values $x > y > 0$ therefore $f(x) \geq f(y)$ due to the function being increasing. $-x < -y < 0$ is a pretty obvious deduction => $f(-x) \leq f(-y)$. But the function is even therefore: $f(-x) = f(x)$ and $f(-y) = f(y)$. Due to the inequality mentioned a bit earlier, we would have: $f(x) \geq f(y)$ and $f(x) \leq f(y)$ therefore $f(x) = f(y)$.
Am I on the right track? These were my ideas so far.
There is not need to mention $0$. Suppose that $f$ is increasing. If $f$ is not constant, then there are two numbers $x$ and $y$ with $x\ne y$ and $f(x)<f(y)$. Suppose that $x<y$. Then $-y<-x$ and$$f(-y)=f(y)>f(x)=f(-x).$$But this is impossible, since $f$ is increasing and $-y<-x$. If $x>y$ or if $f$ is decreasing, the argument is similar.