$x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2}+ x\dfrac{\partial u}{\partial x}+ y \dfrac{\partial u}{\partial y} \ \; \ \; \ For \; u= e^{x+y} \; + \; log(x^3+y^3-x^2y-xy^2) \ \; \ $

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

$ \therefore z \;=\; x^n \phi \bigg( \dfrac{y}{x} \bigg) ......(i).... $

{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= e^{x+y} + log(x^3+y^3-x^2y-xy^2) \; \; say, \; v+w \\ \; \\ v \;=\; e^{x+y} $

Replace $x$ by $tx$ and $y$ by $+y$ , we get $ v \;=\; e^{t(x+y)}$ which is not a homogeneous function But log v= x+y is a homogeneous function $log v\;=\; t^1 (x+y) logv $ is a homogeneous function with degree n=1 $ \therefore f(v) \;=\; log v $ $ \therefore $ By corollary of Euler’s theorem, $ x \dfrac{\partial v}{\partial x} + y \dfrac{\partial v}{\partial y} \;=\; n \dfrac{f(v)}{f'(v)} \;=\; 1 \cdot \dfrac{log \; v}{ \frac{1}{v}} \;=\; vlogv \;=\; e^{x+y} (x+y) \ \; \ $ Also, $ x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} \ \;=\; G(v)\bigg[ G'(v)-1 \bigg] \ \; \ G(v) \;=\; \dfrac{nf(v)}{f^1(v)} \;=\; vlogv \ \; \ \therefore G'(v) \;=\; v \Big( \frac{1}{v} \Big) + logv(1) \;=\; 1+logv $ $ \therefore x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} \;=\; \ vlogv[1+logv-1] \;=\; v(logv)^2 \;=\; e^{x+y} (x+y)^2 $ $ w\;=\; log(x^3+y^3-x^2y-xy^2) $ is not a homogeneous function But $ f(w) \;=\; e^w \;=\; x^3+y^3-x^2y-xy^2 $ is a homogeneous function with degree n=3 $ \Big[ e^w \;=\; (tx)^3+(ty)^3-(tx)^2ty-tx(ty)^2 \;=\; t^3( x^3+y^3-x^2y-xy^2 )\Big] $ $\therefore$ By corollary of Euler’s theorem $ x \dfrac{\partial v}{\partial x} + y \dfrac{\partial v}{\partial y} \;=\; \;=\; n \dfrac{f(w)}{f'(w)} \;=\; 3 \cdot \dfrac{e^w}{e^w} \;=\; 3 $ Also, $ x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} \;=\; G(w)[G'(w)-1] $ Where $ G(w) \;=\; \dfrac{f(w)}{f'(w)} \;=\; 3 \; \; \therefore G'(w)=0 $ $ x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} \;=\; 3(0-1) \;=\; -3 $ $ \because u=v+w$ $ \therefore x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} \;=\; x^2 \dfrac{\partial^2 v}{\partial x^2}+ \ 2xy\dfrac{\partial^2 v}{\partial x \partial y}+ y^2\dfrac{\partial^2 v}{\partial y^2} + x^2 \dfrac{\partial^2 w}{\partial x^2}+ 2xy\dfrac{\partial^2 w}{\partial x \partial y}+ y^2\dfrac{\partial^2 w}{\partial y^2}
$ Also $ \ x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} \;=\; x \dfrac{\partial v}{\partial x} + y \dfrac{\partial v}{\partial y} + x \dfrac{\partial w}{\partial x} + y \dfrac{\partial w}{\partial y} \ \; \ $ Adding them, $ \therefore x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2} + x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} \;=\; \ e^{x+y} (x+y)^2 -3 + e^{x+y}(x+y) \;=\; e^{x+y} (x+y)[1+(x+y)] $