Given dy / DX = Y / 2x-x / 2Y (y = 0 when x = a), the original equation is solved

x+(y2/x)-a=0

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1. Derived from (DX) / (Dy) = 1 / y ': (d ^ 2x) / (Dy ^ 2) = - y' '/ (y') ^ 3 rt In addition, what does "d" mean here? Tongji 6 p.90 suddenly comes out with the derivative function expressed by D, and the concept of "d" has not been explicitly put forward What about (d ^ 3) x / D (y ^ 3)
2. Finding the solution of dy / DX = 1 / (2x-y ^ 2)
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17. Is there a real number pair (x, y) so that SiNx = cosy and arcsinx = arccosy hold simultaneously? Why?
18. If x and y are real numbers and SiNx * cosy = 1, then cos (x + y)=
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