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由y-xe^(y-1)=1得，y(0)=1
y'-e^(y-1)-xe^(y-1) *y'=0，所以，y'(0)=1
y"-e^(y-1) *y' -e^(y-1) *y' -xe^(y-1) *(y')^2 -xe^(y-1) *y''=0，所以，y"(0)=2
dz/dx = f'(lny-sinx) * ((1/y)*y' -cosx)
dz/dx |(x=0) = f'(lny-sinx) * ((1/y)*y' -cosx) |(x=0) = 1*0 =0
d^2z/dx^2 |(x=0) = f"(lny-sinx) *((1/y)*y' -cosx)^2 +f'(lny-sinx) *(-(1/y^2)*y' +(1/y)*y" +sinx) |(x=0)
=0 +1
=1
y'-e^(y-1)-xe^(y-1) *y'=0，所以，y'(0)=1
y"-e^(y-1) *y' -e^(y-1) *y' -xe^(y-1) *(y')^2 -xe^(y-1) *y''=0，所以，y"(0)=2
dz/dx = f'(lny-sinx) * ((1/y)*y' -cosx)
dz/dx |(x=0) = f'(lny-sinx) * ((1/y)*y' -cosx) |(x=0) = 1*0 =0
d^2z/dx^2 |(x=0) = f"(lny-sinx) *((1/y)*y' -cosx)^2 +f'(lny-sinx) *(-(1/y^2)*y' +(1/y)*y" +sinx) |(x=0)
=0 +1
=1
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