Solution:
If one of the numbers $a, b$ and $c$ is equal to zero, it is easy to see that the other two numbers also have to be equal to zero, which gives us the solution $(0,0,0)$.
Now assume that $a, b, c \neq 0$.
If all three numbers are positive, then the left-hand side of each equation is positive, while the right-hand sides are equal to zero, which is impossible.
Let us assume that only one of the numbers is positive, and without loss of generality let it be $a$. Since $b, c<0$, it follows that $b^{2}+b c+a>0$, which is a contradiction.
It remains to consider the two following cases:

a. All three numbers are negative.
We substitute $a=-x, b=-y$ and $c=-z$, where $x, y, z>0$. The original system transforms into
$$\begin{aligned} & x^{2}+x y=z \\ & y^{2}+y z=x \\ & z^{2}+z x=y \end{aligned}$$
The system is cyclic, so we can assume that $x \leqslant y$ and $x \leqslant z$. Now we have
$$\begin{aligned} & x^{2}+x y=z \geqslant x \Longrightarrow x+y \geqslant 1 \\ & y^{2}+y z=x \leqslant y \Longrightarrow y+z \leqslant 1 \end{aligned}$$
From the previous two inequalities we conclude that
$$x+y \geqslant 1 \geqslant y+z, \quad \text{ i.e. } \quad x \geqslant z$$
On the other hand $x \leqslant z$, so we get $x=z$.
Now, from equation (1) it follows that $x+y=1$, while from equation (2) it follows that
$$x=y^{2}+y z=y^{2}+y x=y(y+x)=y$$
Thus $x=y=z$ and from $x+y=1$ we see that $x=y=z=1 / 2$.
We easily verify that $(a, b, c)=\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$ is indeed a solution.

b. Exactly one of the numbers is negative.
Without loss of generality we can assume that $c$ is negative, while $a$ and $b$ are positive. From the second equation we conclude that $b(b+c)=-a<0$, thus $b+c<0$. The third equation yields $c(a+c)=-b<0$, thus $a+c>0$.
Adding $a+b$ to the first equation and cyclic permutation yields
$$a+b+c=(1-a)(a+b)=(1-b)(b+c)=(1-c)(c+a)$$
The last product is positive. This implies that $1-a>0$ and $1-b<0$ by our above considerations. Therefore $0<a+c<1+c<b+c<0$, a contradiction.

Solution by symmetric functions. We set
$$p:=a+b+c, \quad q:=a b+a c+b c, \quad r=a b c$$
Our strategy will be to determine $p, q$ and $r$ by considering equations of the form
$$\sum_{\text{cyc}} f(a, b, c)\left(a^{2}+a b+c\right)=0$$
By setting $f(a, b, c)=1$, we find $p^{2}-q+p=0$.
By setting $f(a, b, c)=b$, we find $p q-3 r+q=0$.
(Here we use the general identity $\sum_{\text{cyc}}\left(a^{2} b+a b^{2}\right)=p q-3 r$.)
By setting $f(a, b, c)=c^{2}$, we find $\left(q^{2}-2 p r\right)+p r+\left(p^{3}-3 p q+3 r\right)=0$.
By elimination of $q=p^{2}+p$ and $r=\frac{p q+q}{3}=\frac{p(p+1)^{2}}{3}$ we find
$$\left(p^{2}+p\right)^{2}-p \cdot \frac{p(p+1)^{2}}{3}+p^{3}-3 p\left(p^{2}+p\right)+p(p+1)^{2}=0$$
which is equivalent to
$$0=2 p^{4}+p^{3}-p^{2}+3 p=p(2 p+3)\left(p^{2}-p+1\right)$$
Now we see that either $p=0$ or $p=-\frac{3}{2}$.
In the case $p=0$ we find that also $q=0$ and $r=0$, whence $a=b=c=0$. In the case $p=-\frac{3}{2}$ we find $q=\frac{3}{4}$ and $r=-\frac{1}{8}$, hence $a, b, c$ are the solutions to the cubic equation
$$0=x^{3}+\frac{3}{2} x^{2}+\frac{3}{4} x+\frac{1}{8}=\left(x+\frac{1}{2}\right)^{3}$$
This gives $a=b=c=-\frac{1}{2}$.

Solution:
As in the first solution, we prove that as soon as one of the variables is 0, all three variables have to be 0. Obviously, the triple $(0,0,0)$ is a solution. From now on, we may therefore assume that all three variables are non-zero.
We can rewrite the system of equations as
$$\begin{aligned} & -c=a(a+b) \\ & -a=b(b+c) \\ & -b=c(c+a) \end{aligned}$$
Multiplying these equations and dividing by $a b c \neq 0$ gives
$$(a+b)(b+c)(c+a)=-1$$
On the other hand by summing up the equations we get
$$-a-b-c=a^{2}+b^{2}+c^{2}+a b+b c+a c$$
Now we substitute
$$x=a+b, \quad y=b+c, \quad z=c+a$$
which transforms equation (4) and (5) into
$$x y z=-1, \quad-\frac{x+y+z}{2}=\frac{x^{2}+y^{2}+z^{2}}{2}$$
Now we calculate
$$3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(|x|+|y|+|z|)^{2} \geqslant|x+y+z|^{2}=\left(x^{2}+y^{2}+z^{2}\right)^{2} \geqslant 3^{2}\{\sqrt[3]{x^{2} y^{2} z^{2}}\}^{2}=9$$
where the first inequality comes from Cauchy-Schwarz (or QM-AM), the second one from the triangle inequality and the last one from AM-GM. The equalities come from (6).
Now, if we denote $S=x^{2}+y^{2}+z^{2}$, we have the inequalities
$$3 S \geqslant S^{2} \geqslant 9$$
and because we trivially have $S>0$ (note that $S=0$ would imply $x=y=z=0$ and hence $a=b=c=0$, which has already been excluded), we can split it into the inequalities $3 \geqslant S$ and $S \geqslant 3$, so we have equality and actually all the inequalities are equalities.
The case of equality for the triangle equality is when all nonzero $x, y, z$ have the same sign and, in view of equations 6, the only possibility is that $x, y, z$ are all negative. Moreover, in the last inequality, we have equality exactly when $x^{2}=y^{2}=z^{2}$ and, because they have the same sign, it means $x=y=z$. Finally, in view of $x y z=-1$, the only possibility is $x=y=z=-1$. By definition of $x, y, z$ the values of $a, b, c$ are then
$$(a, b, c)=\left(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\right)$$
and this is indeed a solution.

The solutions are
$$(a, b, c) \in\left\{(0,0,0),\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)\right\}$$