设函数f与g均可微,z=f(xy,lnx+g(xy)),则x*z关于x的微分-y*z关于y的微分=

设u=xy,v=lnx+g(xy),则 x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：
dz=(∂f/∂u)d(xy)+(∂f/∂v)d(lnx+g(xy))
=(∂f/∂u)[ydx+xdy]+(∂f/∂v)[d(lnx)+d(g(xy))]
=(∂f/∂u)[ydx+xdy]+(∂f/∂v)[(1/x)dx+g'(xy)d(xy)]
=(∂f/∂u)[ydx+xdy]+(∂f/∂v)[(1/x)dx+g'(xy)(ydx+xdy)]
=[y(∂f/∂u)+((1/x)+yg'(xy))(∂f/∂v)]dx+[x(∂f/∂u)+xg'(xy)(∂f/∂v)]dy
则
∂z/∂x=y(∂f/∂u)+((1/x)+yg'(xy))(∂f/∂v)；
∂z/∂y=x(∂f/∂u)+xg'(xy)(∂f/∂v).
故x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.