Start with the given equation and differentiate with respect to x both sides of the equation
#2=y*sin x-x*cos y#
#d/dx(2)=d/dx(y*sin x-x*cos y)#
#0=y*d/dx(sin x)+sin x*d/dx(y)-[x*d/dx(cos y)+cos y*d/dx(x)]#
#0=y*cos x+sin x*y'-[x(-sin y*y')+cos y*1]#
#0=y*cos x+sin x*y'+x*sin y*y'-cos y#
transpose terms without the #y'#
#cos y-y*cos x=sin x*y'+x*sin y*y'#
by symmetric property of equality we have
#sin x*y'+x*sin y*y'=cos y-y*cos x#
factor out #y'#
#(sin x+x*sin y)*y'=cos y-y*cos x#
Divide both sides of the equation by #(sin x+x*sin y)#
#((sin x+x*sin y)*y')/(sin x+x*sin y)=(cos y-y*cos x)/(sin x+x*sin y)#
#(cancel(sin x+x*sin y)*y')/cancel(sin x+x*sin y)=(cos y-y*cos x)/(sin x+x*sin y)#
#y'=(cos y-y*cos x)/(sin x+x*sin y)#
God bless....I hope the explanation is useful.
