Derivada

ln(x/y) = e^(y/x) ---> ln(x) - ln(y) = e^(y/x) ---> ln(y) = ln(x) - e^(y/x)

Derivando y implicitamente, fica:

y'/y=1/x - e^(y/x) (y'x-y)/x²

---> y' = y[1/x - e^(y/x)(y'x-y)/x²]

---> y' = y/x - ye^(y/x) (y'x-y)/x²

---> y' + [ye^(y/x)-y²e^(y/x)]/x² = y/x

---> y'x² + ye^(y/x)y'x - y²e^(y/x)=yx

---> y'(x²+xye^(y/x) = yx+y²e^(y/x)

---> y' = [yx +y²e^(y/x)]/[x²+xye^(y/x)]

Conclusão: para ln(x/y) = e^y/x,      y' = [yx +y²e^(y/x)]/[x²+xye^(y/x)]