Since, $z=f(x,y)$ is a homogeneous function of degree n

$ \therefore z \;=\; x^n \phi \bigg( \dfrac{y}{x} \bigg) \; \; \ldots (i) \ldots $ {By Property of homogeneous function }

Differentiating equation (i) partially w.r.t. x,

$ \therefore \dfrac{\partial z}{\partial x} \;=\; \phi \bigg( \dfrac{y}{x} \bigg) \cdot n x^{n-1} \;+\; x^n \phi' \bigg( \dfrac{y}{x} \bigg) \dfrac{\partial }{\partial x} \bigg( \dfrac{y}{x} \bigg) \\ \; \\ \; \\ \therefore \dfrac{\partial z}{\partial x} \;=\; n x^{n-1} \phi \bigg( \dfrac{y}{x} \bigg) \;+\; x^n \phi' \bigg( \dfrac{y}{x} \bigg) \bigg( \dfrac{-y}{x^2} \bigg) \; \; \ldots (ii) $

Differentiating equation (i) partially w.r.t. y,

$ \dfrac{\partial z}{\partial y} \;=\; x^n \phi' \bigg( \dfrac{y}{x} \bigg) \dfrac{\partial }{\partial y} \bigg( \dfrac{y}{x} \bigg)n \;+\; \phi \bigg( \dfrac{y}{x} \bigg) \dfrac{\partial }{\partial y} (x^n) \\ \; \\ \; \\ \therefore \dfrac{\partial z}{\partial y} \;=\; x^n \phi' \bigg( \dfrac{y}{x} \bigg) \bigg( \dfrac{1}{x} \bigg)n \;+\; \phi \bigg( \dfrac{y}{x} \bigg) (0) \; \; \; \ldots (iii) $

Multiplying equation (ii) by x throughout and equation (iii) by y and adding them, we get

$ x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} \;=\; nx^n\phi \bigg( \dfrac{y}{x} \bigg) + x^{n-2+1} \phi. \bigg( \dfrac{y}{x} \bigg) (-y) + x^{n-1}y \phi' \bigg( \dfrac{y}{x} \bigg) \\ \; \\ \; \\ \; \\ \therefore x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} \;=\; nx^n\phi \bigg( \dfrac{y}{x} \bigg) - x^{n-1}y \phi. \bigg( \dfrac{y}{x} \bigg) + x^{n-1}y \phi' \bigg( \dfrac{y}{x} \bigg) \\ \; \\ \; \\ \; \\ x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} \;=\; nx^n\phi \bigg( \dfrac{y}{x} \bigg) \;=\; nz \;\;\; \ldots From \; (i) $

Hence Proved.

$ \\ \; \\ u=\dfrac{\sqrt{x}+\sqrt{y}}{x+y} \;=\; \dfrac{ \sqrt{x} \bigg(1+ \dfrac{\sqrt{y}}{\sqrt{x}} )}{x(1+\dfrac{y}{x})} \\ \; \\ = x^{-1/2} \phi \bigg( \dfrac{y}{x} ) $

$ \therefore$ u is a homogeneous function with degree $ n= -1/2 $

$ \therefore x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y} \; = \; nu \\ \; \\ \; \\ x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y} \;=\; \dfrac{-1}{2} \bigg( \dfrac{\sqrt{x}+\sqrt{y}}{x+y} \bigg) $