, 2018

Denote ($*$) as the given condition. Replacing $x = y = -\frac{1}{2}$ in ($*$), we get
$$f\left(2\left(f\left(-\frac{1}{2}\right)\right)^{2}-\frac{1}{2}\right)=0$$
If there exists some $a \neq 0$ such that $f(a)=0$. Substituting $x=a$ in ($*$), we get $f(y)=0$ for any $y$, which is obviously a solution.

Let us consider now the case $f(x) \neq 0$ for any nonzero real number $x$. Hence, $f(0)=0$ and $f\left(-\frac{1}{2}\right)= \pm \frac{1}{2}$. On the other hand, it is easy to see that if $f$ is solution, then $-f$ is also a solution. Therefore, we only need to consider the case $f(0)=0$ and $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. From this, we have
$$\begin{aligned} f\left(-\frac{1}{2}-f(y)\right) & =-f(y)-\frac{1}{2}, \quad \forall y \in \mathbb{R} \\ f\left(\frac{1}{2}\right) & =\frac{1}{2} \end{aligned}$$
and
$$f(x-f(x))=f(x)-x, \quad \forall x \in \mathbb{R}$$
Therefore,
$$\begin{aligned} f\left(x-\frac{1}{2}-f(x)\right) & =f\left(-\frac{1}{2}-f(x-f(x))\right) \\ & =-\frac{1}{2}-f(x-f(x))=x-\frac{1}{2}-f(x) \end{aligned}$$
which implies $f\left(-f\left(\frac{1}{2}\right)\right)=-f\left(\frac{1}{2}\right)$. Now, substituting $x=\frac{1}{2}$ and $y= -f\left(\frac{1}{2}\right)$ in ($*$), we get
$$f\left(\frac{1}{2}-2\left(f\left(\frac{1}{2}\right)\right)^{2}\right)=0 .$$
Thus $f\left(\frac{1}{2}\right)= \pm \frac{1}{2}$. If $f\left(\frac{1}{2}\right)=-\frac{1}{2}$, then $\frac{1}{2}=-f\left(\frac{1}{2}\right)=f\left(-f\left(\frac{1}{2}\right)\right)=f\left(\frac{1}{2}\right)= -\frac{1}{2}$, contradiction. Therefore, $f\left(\frac{1}{2}\right)=\frac{1}{2}$.

From here, it follows that
$$f\left(f(y)+\frac{1}{2}\right)=f(y)+\frac{1}{2}, \quad \forall y \in \mathbb{R}$$
In this equation, substituting $y=x-\frac{1}{2}-f(x)$, we get
$$f(x-f(x))=x-f(x), \quad \forall x \in \mathbb{R}$$
Comparing this result with $f(x-f(x))=f(x)-x$, we get $f(x)=x$ for any $x$, which is clearly a solution.

In conclusion, there are three solutions: $f(x)=0$, $f(x)=x$ and $f(x)= -x$ for all real number $x$.