9. The answer is (B). Subtracting the equations member by member, we get

$$
(y-x)(-1+x+y)=(y-x)\left(y^{2}+x y+x^{2}\right) \text {, }
$$

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 text, we get $x+x^{2}=x^{3}$, 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 $x \neq y$, it is necessary to have $x^{2}+y^{2}+x y-x-y+1=0$. 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). Therefore, there are exactly three solutions, all with $x=y$.
9. How many pairs of real numbers $(x, y)$ satisfy both equations $x+y^{2}=y^{3}$ and $y+x^{2}=x^{3}$?
(A) 1
(B) 3
(C) 5
(D) 9
(E) Infinite