Given equation
#-1=yx+(x-ye^x)/(y+x)#
Differentiate with respect to x both sides of the equation
#-1=yx+(x-ye^x)/(y+x)#
#d/dx(-1)=d/dx(yx)+d/dx((x-ye^x)/(y+x))#
#0=y*d/dx(x)+x*d/dx(y)+((y+x)d/dx(x-ye^x)-(x-ye^x)*d/dx(y+x))/(y+x)^2#
#0=y*1+x*y'+((y+x)(1-(y*d/dx(e^x)+e^x*d/dx(y)))-(x-ye^x)*(y'+1))/(y+x)^2#
#0=y+x*y'+((y-y^2*e^x-e^x*yy'+x-xy*e^x-xe^x*y'-xy'+ye^xy'-x+ye^x))/(y+x)^2#
#0=y+xy'+((y-y^2e^x-xye^x-xe^xy'-xy'+ye^x))/(y+x)^2#
#0=y(y+x)^2+x(y+x)^2y'+y-y^2e^x-xye^x-xe^xy'-xy'+ye^x#
#xye^x+y^2e^x-y(y+x)^2-y-ye^x=x(y+x)^2y'-xe^xy'-xy'#
#y'=(xye^x+y^2e^x-y(y+x)^2-y-ye^x)/(x(y+x)^2-xe^x-x)#
God bless...I hope the explanation is useful.
