Note that $a=f(f(f(a)))$ where $f(x)=\frac{2x}{1+x^2}$. Fixed points: $\frac{2x}{1+x^2}=x\implies x^3-x=0\implies x(x+1)(x-1)=0\implies x\in\{-1,0,1\}$. Three-cycle points: Since $\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$, we have $f(i\tan\theta)=i\tan 2\theta$ and so we need $\tan 8\theta=\tan\theta\implies\theta=\frac{\pi k}{7}$ for $k\in\mathbb{Z}$. This gives the following results: Real: $(-1,-1,-1)$, $(0,0,0)$, $(1,1,1)$ Complex: $\left(i\tan\frac{\pi}{7},i\tan\frac{2\pi}{7},i\tan\frac{4\pi}{7}\right)$, $\left(i\tan\frac{2\pi}{7},i\tan\frac{4\pi}{7},i\tan\frac{\pi}{7}\right)$, $\left(i\tan\frac{4\pi}{7},i\tan\frac{\pi}{7},i\tan\frac{2\pi}{7}\right)$ Only three real solutions!

So, here is the solution by the author. It is obvious that the triplets $(0, 0, 0), (1, 1, 1), (-1, -1, -1)$ satisfy the given system. Now, we will prove that these are the only solutions. Firstly, one can easily verify that if one of $a, b, c$ is zero, then all are zero. Thus, we look for non-zero solutions. We can rewrite the first equation as $$ \frac{2a}{b} = a^2 + 1$$implying that $a$ and $b$ have the same sign. Similarly, we can deduce that $b$ and $c$, $c$ and $a$ must have the same sign. Thus, $a, b, c$ are either all positive or all negative. W.L.O.G we assume the first case; otherwise, we can multiply each of the equations by $-1$. Adding the three original equations gives $$ a + b + c = a^2 b + b^2 c + c^2 a.$$Moreover, we can rewrite the given equations as $$\frac{2}{b}-\frac{1}{a} = a, \qquad \frac{2}{c} - \frac{1}{b} = b, \qquad \frac{2}{a} - \frac{1}{c} = c.$$Adding these equations gives $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = a+b+c $$To recall, we have $$a^2 b +b^2 c +c^2a = a + b+ c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$and let us denote this value by $S$. Now, we can safely apply Cauchy's Inequality so that $$S^2 = (a^2 b+b^2c+c^2a)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right) \geq (a+b+c)^2 = S^2 .$$Thus, we must have $$a^2 b^2 = b^2 c^2 = c^2 a^2 \implies a=b=c$$and substituting back in the given system gives $a=b=c=1$.