Three. (25 points) Find the integer solutions of the indeterminate equation
$$
9 x^{2}-6 x-4 y^{4}+12 y^{3}-12 y^{2}+5 y+3=0
$$
Three, rearranging and partially completing the square of the original equation, we get
$$
(3 x-1)^{2}=\left(2 y^{2}-3 y\right)^{2}+3 y^{2}-5 y-2 \text {, }
$$

and $(3 x-1)^{2}=\left(2 y^{2}-3 y+1\right)^{2}-\left(y^{2}-y+3\right)$.
If the integer $y \neq 0,1,2$, then it must be that
$$
2 y^{2}-3 y, 2 y^{2}-3 y+1,3 y^{2}-5 y-2, y^{2}-y+3
$$

are all positive integers simultaneously.
Furthermore, the perfect square $(3 x-1)^{2}$ must lie between two consecutive perfect squares $\left(2 y^{2}-3 y\right)^{2}$ and $\left(2 y^{2}-3 y+1\right)^{2}$, which is a contradiction.
Therefore, it is only necessary to check $y=0,1,2$.
When $y=0$, $(3 x-1)^{2}=-2$, no solution;
When $y=1$, $(3 x-1)^{2}=-3$, no solution;
When $y=2$, $(3 x-1)^{2}=4$, solving gives $x=1$, which satisfies the requirement.
In summary, the solution to the original indeterminate equation is $x=1, y=2$.