图中做标记的两个题，要详细过程谢谢
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（1）两边分别求导：
y'=cos(x+y).(1+y')
y'/(1+y')=cos(x+y)
y'=cos(x+y)/[1-cos(x+y)]
y'=cos(x+y).(1+y')两边再次求导
y"=-sin(x+y)(1+y')²+cos(x+y)y"
y"=-sin(x+y)(1+y')²/[1-cos(x+y)]
=-sin(x+y)(1+cos(x+y)/[1-cos(x+y)])²/[1-cos(x+y)]
=-sin(x+y)(1/[1-cos(x+y)])²/[1-cos(x+y)]
=-sin(x+y)/[1-cos(x+y)]³
=-y/[1-cos(x+y)]³
y'=cos(x+y).(1+y')
y'/(1+y')=cos(x+y)
y'=cos(x+y)/[1-cos(x+y)]
y'=cos(x+y).(1+y')两边再次求导
y"=-sin(x+y)(1+y')²+cos(x+y)y"
y"=-sin(x+y)(1+y')²/[1-cos(x+y)]
=-sin(x+y)(1+cos(x+y)/[1-cos(x+y)])²/[1-cos(x+y)]
=-sin(x+y)(1/[1-cos(x+y)])²/[1-cos(x+y)]
=-sin(x+y)/[1-cos(x+y)]³
=-y/[1-cos(x+y)]³
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（2）两边求导
y²十2xyy'-cos（πy²）×2πyy'=0
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