Three, Solution Using the substitution method, we get
$$x=\frac{x^{4}}{4(x-1)^{3}+4(x-1)} \text {. }$$

Rearranging gives
$$x(x-2)\left(3 x^{2}-6 x+4\right)=0 \text {, }$$

Solving this yields
$$\begin{array}{l} x_{1}=0, \\ x_{2}=2, \\ x_{3}=1+\frac{\sqrt{3}}{3} i, \\ x_{4}=1-\frac{\sqrt{3}}{3} i . \end{array}$$

Substituting into $y$ gives
$$\begin{array}{l} y_{1}=0, \\ y_{2}=2, \\ y_{3}=1-\frac{\sqrt{3}}{3} i, \\ y_{4}=1+\frac{\sqrt{3}}{3} i . \end{array}$$

Thus, there are 4 sets of solutions
$$\begin{array}{l} \left\{\begin{array}{l} x_{1}=0, \\ y_{1}=0 ; \end{array}\right. \\ \left\{\begin{array}{l} x_{2}=2, \\ y_{2}=2, \end{array}\right. \\ \left\{\begin{array}{l} x_{3}=1+\frac{\sqrt{3}}{3} i, \\ y_{3}=1-\frac{\sqrt{3}}{3} i ; \end{array}\right. \\ \left\{\begin{array}{l} x_{4}=1-\frac{\sqrt{3}}{3} i, \\ y_{4}=1+\frac{\sqrt{3}}{3} i . \end{array}\right. \end{array}$$