两边对x求导：y+xy'-cos(πy2) (2πyy')=0      y'[x-2πycos(πy^2)]=-y

                      y' = 2πcos(πy^2) - x/y                  (1)

2xy+x^2 y'+ 2yy'cos(y^2)=0          y'[x^2+2ycos(y^2)]=-2xy

                      y' = -2xy / [x^2+2ycos(y^2)]           (2)