Assume that f is differentiable and z=f(x/y).Show that x(dz/dx) +y(dz/dy)?

Answer 1

z=f(x/y)
Let x/y =w
dz/dx = (dz/dw ) * (dw/dx) Chain rule
= (df/dw) * (1/y)
dz/dy = (df/dw) * (dw/dy)
= (df/dw) * (-x/y^2)

Hence,
x*(dz/dx) + y*(dz/dy)
= (x/y) * (df/dw) - (x/y) * (df/dw)
=0

Cheers.

Answer 2

I think you're asking to show that x(dz/dx)+y(dz/dy)=0

dz/dx=f'(x/y)*1/y
dz/dy=-f'(x/y)*x/y^2

x(dz/dx)+y(dz/dy)=f'(x/y)*x/y-f'(x/y)*(xy)/y^2
=f'(x/y)*x/y-f'(x/y)*x/y=0 QED