谁对谁求导？？题问不清
y'=(x^1/y)'
=x^(1/y)*lnx*(1/y)'+(1/y)*x^(1/y-1)
=y*lnx*(-y'/y^2)+x^(1/y)/(xy)
=-y'lnx/y+1/x
从而
y'(1+lnx/y)=1/x
y'=y/[x(lnx+y)]
y''={y'*[x(lnx+y)]-y*[x(lnx+y)]'}/[x(lnx+y)]^2
={y-y*[lnx+y+1+xy')]}/[x(lnx+y)]^2
=-y*[lnx+y+y/(lnx+y)]/[x(lnx+y)]^2
=-y*[(lnx+y)^2+y]/[x^2(lnx+y)^3]
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