Prerequisites : The Usual Rules of Differentiation.
Given that, e^(x/y)=x-yexy=x−y.
:. lne^(x/y)=ln(x-y).
:. x/y=ln(x-y), or, x=yln(x-y).
:. d/dx{x}=d/dx{xln(x-y)}.
:.1=xd/dx{ln(x-y)}+ln(x-y)*d/dx{y},
i.e., 1=x*1/(x-y)*d/dx{x-y}+ln(x-y)*dy/dx,
or, 1=x/(x-y){d/dx(x)-d/dx(y)}+ln(x-y)*dy/dx.
:. 1=x/(x-y){1-dy/dx}+ln(x-y)*dy/dx.
:. 1=x/(x-y)-x/(x-y)*dy/dx+ln(x-y)dy/dx.
:. 1-x/(x-y)={ln(x-y)-x/(x-y)}dy/dx.
:. {(x-y)-x}/cancel(x-y)={(x-y)ln(x-y)-x}/cancel(x-y)*dy/dx.
:. -y={(x-y)ln(x-y)-x}dy/dx.
rArr dy/dx={x-(x-y)ln(x-y)}/y.
Since, (x-y)=e^(x/y) and ln(x-y)=x/y, we may write,
dy/dx=x/y-1/y*e^(x/y)*x/y=x/y{1-1/y*e^(x/y)}.