高等数学求偏导数这题怎么做： x z=ln tan----- y题目是z=ln tan x/y希望能把这道题的步骤给我看

dz =[1/tan(x/y)]*sec^2(x/y) *[dx/y + (-x/y^2)dy]
dz/dx = 2 /[y *sin(2x/y)]
dz/dy= - 2x/[y^2*sin(2x/y)]
dz/dx =d (ln tan x/y）/dx
=(1/tan( x/y)) *d (tan x/y)/dx
=(1/tan( x/y)) *sec^2( x/y)*d(x/y)
=(1/tan( x/y)) *sec^2( x/y)*dx/y
=2 /[y *sin(2x/y)]
dz/dy =d (ln tan x/y）/dy
=(1/tan( x/y)) *d (tan x/y)/dy
=(1/tan( x/y)) *sec^2( x/y)*d(x/y)
=(1/tan( x/y)) *sec^2( x/y)*(-x/y^2)
= - 2x/[y^2*sin(2x/y)]