$\ 解：(1) 在y=-2x + 6中，令y = 0，则-2x + 6 = 0，解得x = 3；$
$令x = 0，则y = 6.\ \therefore A(3,0)，B(0,6).\ $
$把A(3,0)，B(0,6)代入y=-x^{2}+bx + c，$
$得\begin{cases}-9 + 3b + c = 0\\c = 6\end{cases}，解得\begin{cases}b = 1\\c = 6\end{cases}，$
$\therefore抛物线对应的函数表达式为y=-x^{2}+x + 6$
$\ (2) 存在 设D(t,-t^{2}+t + 6)(0＜t＜3)，则E(t,-2t + 6)，C(t,0).\ $
$\therefore CE=-2t + 6，AC = 3 - t，DE=-t^{2}+3t.$
$\ \because\triangle BDE和\triangle ACE相似，\angle BED=\angle AEC，$
$\therefore\triangle ACE\sim\triangle BDE或\triangle ACE\sim\triangle DBE.\ $
$①如图①，当\triangle ACE\sim\triangle BDE时，\angle ACE=\angle BDE = 90^{\circ}，$
$\therefore BD// AC. \therefore点D的纵坐标为6.\ $
$\therefore -t^{2}+t + 6 = 6，解得t = 0（不合题意，舍去）或t = 1.\ $
$\therefore D(1,6).\ $
$②如图②，当\triangle ACE\sim\triangle DBE时，\angle CAE=\angle BDE.\ $
$过点B作BH\perp CD于点H，则H(t,6)，\angle BHD = 90^{\circ}.$
$\ \therefore BH = t，DH=-t^{2}+t.\ $
$\because\tan\angle BDE=\frac{BH}{DH}，\tan\angle CAE=\frac{OB}{OA}，$
$且\angle BDE=\angle CAE，$
$\therefore\frac{t}{-t^{2}+t}=\frac{6}{3}，解得t = 0（不合题意，舍去）或t=\frac{1}{2}.\ $
$经检验，t=\frac{1}{2}是原方程的解，且符合题意.$
$\ \therefore D(\frac{1}{2},\frac{25}{4}).\ $
$综上所述，点D的坐标为(1,6)或(\frac{1}{2},\frac{25}{4})\ $
$(3) 如图③，\because四边形EGFD为菱形，$
$\therefore DE// FG，DE = FG，DE = EG.\ $
$设D(m,-m^{2}+m + 6)(0＜m＜3)，则E(m,-2m + 6).$
$\ \therefore DE=-m^{2}+3m.\ $
$设F(n,-n^{2}+n + 6)(0＜n＜3，且n\neq m)，则G(n,-2n + 6).\ $
$\therefore FG=-n^{2}+3n.\ \because DE = FG，$
$\therefore -m^{2}+3m=-n^{2}+3n，即(m - n)(m + n - 3)=0.\ $
$\because n\neq m，\therefore m - n\neq_{0}.\ $
$\therefore m + n - 3 = 0，即n = 3 - m.\ \because A(3,0)，B(0,6)，$
$\therefore AO = 3，BO = 6.$
$\ \therefore AB=\sqrt{AO^{2}+BO^{2}}=3\sqrt{5}.\ $
$过点G作GK\perp DE于点K，\therefore KG// AC.$
$\ \therefore\angle EGK=\angle BAC.\ $
$\because\cos\angle EGK=\frac{KG}{EG}，\cos\angle BAC=\frac{AO}{AB}，$
$\ \therefore\frac{n - m}{EG}=\frac{3}{3\sqrt{5}}.\ $
$\therefore EG=\sqrt{5}(n - m)=\sqrt{5}(3 - 2m).\ $
$\because DE = EG，\therefore -m^{2}+3m=\sqrt{5}(3 - 2m).\ $
$\therefore m^{2}-(3 + 2\sqrt{5})m+3\sqrt{5}=0，$
$解得m=\frac{3 + 2\sqrt{5}+\sqrt{29}}{2}（不合题意，舍去）或m=\frac{3 + 2\sqrt{5}-\sqrt{29}}{2}.$
$\ \therefore点D的横坐标为\frac{3 + 2\sqrt{5}-\sqrt{29}}{2}$
