解：?$(1)$?根據(jù)題意，
?$(x，kx)*(y， - y)=x^2+(-y)^2-kx·y$?
?$=x^2-kxy + y^2$?
?$ $?因為?$(x，kx)*(y， - y)$?是一個完全平方式，
所以?$-kxy=\pm 2xy，$?解得?$k = \pm 2。$?
?$ (2)$?根據(jù)題意，
?$(3x + y，2x^2+3y^2)*(3，x - 3y)$?
?$=(3x + y)^2+(x - 3y)^2-3(2x^2+3y^2)$?
?$ =(9x^2+6xy + y^2)+(x^2-6xy + 9y^2)-(6x^2+9y^2)$?
?$=4x^2+y^2=84。$?
?$ $?因為?$2x + y = 10，$?
所以?$(2x + y)^2=4x^2+4xy + y^2=100，$?
所以?$4xy=100 - 84 = 16，$?
所以?$xy = 4。$?
?$ (3)$?由?$ (2)$?可知，?$2x + y = 10，$??$xy = 4。$?
?$ $?因為四邊形?$ABCD$?和四邊形?$CEFG $?均為長方形，
所以?$CD = AB = 2x，$??$BC = AD = 8x，$?
?$CG = EF = 4y，$??$CE = FG = y，$?
所以?$DE = 2x - y，$??$BG = 8x - 4y。$?
所以陰影部分的面積為
?$S = S_{\triangle BCD}-S_{\triangle BFG}-S_{\triangle DEF}-S_{\triangle CEG}$?
?$=\frac {1}{2}BC·CD-\frac {1}{2}BG·FG-\frac {1}{2}EF·DE-\frac {1}{2}CG·CE$?
?$ =\frac {1}{2}×8x·2x-\frac {1}{2}(8x - 4y)·y-\frac {1}{2}×4y(2x - y)-\frac {1}{2}×4y·y$?
?$ =8x^2-4xy + 2y^2-4xy + 2y^2-2y^2$?
?$=8x^2-8xy + 2y^2$?
?$=2(4x^2+4xy + y^2)-16xy$?
?$ $?因為?$(2x + y)^2=4x^2+4xy + y^2，$?
所以原式?$=2×10^2-16×4 = 136。$?
