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Are there any functions of that type? IE. idempotent but not on composition, through function multiplication instead.

Sorry, I forgot to mention this, but $f(x)\not\equiv 1$ and $f(x) \not\equiv 0.$

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    Such functions have a simple characterization: the set of characteristic functions of subsets of $\mathbb{R}$. – 2017-02-08
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    @MathematicsStudent1122: Assuming that $f$ is a real-valued function on a subset of $\Bbb R,$ that's true. – 2017-02-08
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    I have adjusted the notation to reflect [the comments](http://math.stackexchange.com/questions/2134170/function-such-that-fx-fx2/2134193#comment4389267_2134193). If the edits are inaccurate, please let me know, and I'll try to fix it. – 2017-02-08

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Your post is a bit ambiguous, so I'll do the best I can.

It seems that you have a function $f:E\to R',$ where $E\subseteq R,$ given some rings $R$ and $R'.$ The following are equivalent for an functions on any ring (each is assumed to hold for all $x\in E$) $$f(x)=\bigl(f(x)\bigr)^2\\0_{R'}=\bigl(f(x)\bigr)^2-f(x)\\0_{R'}=f(x)\cdot f(x)-f(x)\cdot1_{R'}\\0_{R'}=f(x)\cdot\bigl(f(x)-1_{R'}\bigr)$$

Now, if the ring $R'$ has the zero-product property--that is, if $R'$ is a domain--we immediately conclude that $f:E\to\{0_{R'},1_{R'}\}.$ Since $f$ is not identically equal to $0_{R'}$ and not identically equal to $1_{R'},$ we can rule out two functions of this sort. Consequently, there are $2^{|E|}-2$ functions satisfying the desired equation.

If $R'$ is not a domain, then there may be many more functions $f$ that do the trick, but without knowing what $E,$ $R,$ and $R'$ are, this is impossible to say.

If $f$ is never equal to $0_{R'}$ or $1_{R'},$ then $R'$ cannot be a domain, and we require that $f(x)$ and $f(x)-1_{R'}$ must always be zero divisors such that $0_{R'}=f(x)\cdot\bigl(f(x)-1_{R'}\bigr).$ Again, without knowing what $E,$ $R,$ and $R'$ are, it is impossible to say how many such functions there are.

Added: Since your function is apparently supposed to be from [a subset of] the reals into the reals, and is not supposed to be identically zero or identically one, then things are narrowed down quite a bit.

Since $f$ is not supposed to be identically $0$ or identically $1$ on its domain, then we need $E$ to have at least two elements. Now, take any such $E,$ and any $y,z\in E$ with $y\neq z.$ Take any $f:E\to\{0,1\}$ such that $f(y)=0$ and $f(z)=1.$ One can readily show that $0=f(x)\cdot\bigl(f(x)-1\bigr)$ for all $x\in E,$ though $f$ is neither identically $0$ nor identically $1,$ so $f$ satisfies the desired properties.

If we want $f:\Bbb R\to\Bbb R,$ then the set of desired functions $f$ are the indicator functions of all non-empty proper subsets of $\Bbb R.$

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    This question essentially comes from me trying to identify the answer to this question: Show that the ring X = {functions from $\mathbb{R}$ to $\mathbb{R}$} contains the element f such that f $\neq$ 0 f $\neq$ 1 and f = f^2 @CameronBuie – 2017-02-08
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    I have adjusted my answer accordingly. – 2017-02-08
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Where $f(x) \neq 0$ you get $f(x)=1$ after dividing the sides by $f(x).$ So $f(x) \in \{0,1\}.$ And any function of this type satisfies your condition.

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    That assumes that division makes sense. Based on the tags (and the ambiguity of the edit), I'm not sure that's a safe assumption. – 2017-02-08
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If $f(x) = (f(x))^{2}$ for some $x$, then we necessarily have $f(x) = 0$ or $f(x) = 1$, since \begin{split} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f(x) = (f(x))^{2} \\ &\implies f^{2}(x) - f(x) = 0 \\ &\implies f(x)(f(x) - 1) = 0 \\ &\implies f(x) = 0 \text{ or } f(x) = 1 \end{split}

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Well, there are none since solving for f(x) gives f(x)=0 or f(x)=1 (as coffeemath says).

Unless you want functions that are 0 sometimes and 1 sometimes. That could work if you don't need anything continuous.