`z = (xy)/(x-y)` . This is a function of both x and y which are independent variables.

Let's first differentiate z wrt x.

`(dz)/(dx) = (y(x-y)-xy)/(x-y)^2` (x is a constant in this case).

`x(dz)/(dx) = (xy(x-y)-x^2y)/(x-y)^2`

Now let's differentiate z wrt y.

`(dz)/(dy) = (x(x-y)-xy(-1))/(x-y)^2`

`(dz)/(dy) = (x(x-y)+xy)/(x-y)^2`

`y(dz)/(dy) = (xy(x-y)+xy^2)/(x-y)^2`

Now LHS of our expression,

LHS =

`= x(dz)/(dx) + y(dz)/(dy) `

`= (xy(x-y)-x^2y)/(x-y)^2 + (xy(x-y)+xy^2)/(x-y)^2`

`= (xy(x-y)-x^2y+xy(x-y)+xy^2 )/(x-y)^2`

`= (2xy(x-y)+xy^2-x^2y)/(x-y)^2`

`= (2xy(x-y)+xy(y-x))/(x-y)^2`

`= (2xy-xy)/(x-y)`

`= (xy)/(x-y)`

`RHS = z = (xy)/(x-y)`

Therefore LHS = RHS, then

`x(dz)/(dx) + y(dz)/(dy) = z` **Proved.**