System of polynomial equations.

The given system is equivalent to the polynomial equation $z^3-3cz^2+4c^2z-2c^3=0,$ where $z=x+iy.$ Since the equation has the solutions $c, c+c i$ and $c-ci,$ then the corresponding solutions of the system are the pairs $(c, 0),$ $(c, c)$ and $(c, -c),$ and, therefore, the answer is 2.

The equations: $\begin{array}{l} c\neq 0 \\ y\neq 0 \\ -2 c^3+4 x c^2-3 x^2 c+3 y^2 c+x^3-3 x y^2=0 \\ -y^3+4 c^2 y+3 x^2 y-6 c x y=0 \\ \end{array} \Longrightarrow \begin{array}{ll} \text{Over}\ \mathbb{C} & \left[\begin{array}{ll} x\to c & y\to -c \\ x\to c &y\to c\ \\ x\to \left(1-\frac{i}{2}\right) c & y\to -\frac{c}{2} \\ x\to \left(1-\frac{i}{2}\right) c & y\to \frac{c}{2} \\ x\to \left(1+\frac{i}{2}\right) c & y\to -\frac{c}{2} \\ x\to \left(1+\frac{i}{2}\right) c &y\to \frac{c}{2} \\ \end{array}\right] \\ \text{Over}\ \mathbb{R} & \left[ \begin{array}{ll} x\to c & y\to -\sqrt{c^2} \\ x\to c & y\to \sqrt{c^2} \\ \end{array} \right] \\ \end{array}$

In either case, there are 2 real solutions.