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

$ \therefore n= \dfrac{1}{2} \\ \; \\ \; \\ x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y} \;=\; nu $ ...... By Euler's theorem

i.e. $ x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y} \;=\; \dfrac{1}{2} \Bigg( u=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}} \Bigg) $