x=0

0 +e^(-y(0)) = 1

y(0)=0

(0,0)

ysinx+e^(x-y) = 1

Bilateral derivative

ycosx+sinx . y ' +( 1-y')e^(x-y)= 0

(e^(x-y)-+e^(x-y)

y ' =[ycosx +e^(x-y)]/[e^(x-y]-sinx]

y'|(x，y)=(0，0)

=[0 +e^0]/[e^0 -0]

= 1