如图所示，直接求导，把y当做复合函数就行。
(3) y = cos(x+y), y' = -sin(x+y) (1+y'), 得 y' = -sin(x+y)/[1+sin(x+y)]
(4) x^(2/3) + y^(2/3) = a^(2/3), (2/3)x^(-1/3) + (2/3)y^(-1/3)y' = 0,
得 y' = -x^(-1/3)/y^(-1/3) = - (y/x)^(1/3)
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