The answer is (B). Subtracting the equations from each other we obtain
$$(y - x)(-1 + x + y) = (y - x)\left(y^{2} + x y + x^{2}\right),$$
from which we see that at least one of the two equalities $x = y$ and $x^{2} + y^{2} + x y - x - y + 1 = 0$ must hold. If $x = y$, substituting into the equations given in the statement we obtain $x + x^{2} = x^{3}$, from which we easily find the three solutions $(x, y) = (0, 0), \left(\frac{1 + \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right), \left(\frac{1 - \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}\right)$. If instead $x \neq y$ then necessarily $x^{2} + y^{2} + x y - x - y + 1 = 0$ must hold. This last equation can be rewritten as $\frac{1}{2}\left((x + y)^{2} + (x - 1)^{2} + (y - 1)^{2}\right) = 0$, which clearly has no solutions (since at least one of the three numbers $x + y, x - 1, y - 1$ is different from zero). There are therefore exactly three solutions, all with $x = y$.