#x+siny=xy rArr siny=xy-x=x(y-1) rArr x=siny/(y-1)#
#rArr d/dy(x)=d/dy(siny/(y-1))#
#rArr dx/dy={(y-1)d/dy(siny)-sinyd/dy(y-1)}/(y-1)^2#
#={(y-1)cosy-siny}/(y-1)^2#
Since, #siny=x(y-1)#, we have,
#dx/dy={(y-1)cosy-x(y-1)}/(y-1)^2=(cosy-x)/(y-1)#
#:. dy/dx=(y-1)/(cosy-x).......................................................(star)#
#rArr y'(cosy-x)=y-1#
#:. d/dx{y'(cosy-x)}=d/dx(y-1)#
#:. y'd/dx(cosy-x)+(cosy-x)d/dx(y')=dy/dx-0=y'#.
#:. y'{d/dxcos y-d/dx(x)}+(cosy-x)y''=y'#.
#:. y'{(-siny)y'-1}+(cosy-x)y''=y'#.
#:. (cosy-x)y''=y'(y'siny+1)+y'=y'(y'siny+2)#.
#=(y-1)/(cosy-x){((y-1)siny)/(cosy-x)+2}.....................[by (star)]#
#:. y''=(y-1)/(cosy-x)^2{((y-1)siny)/(cosy-x)+2}#
#rArr (d^2y)/dx^2=y''=[(y-1){(y-1)siny+2(cosy-x)}]/(cosy-x)^3#.
Enjoy maths.!
