$Let \ p=\dfrac {x-y}{xy}, q = \dfrac {z-x}{xz} $ $\dfrac {\partial p}{\partial x}= \dfrac {1(x)- (x-y)\times 1}{x^2y} = \dfrac {y}{x^2y}= \dfrac {1}{x^2}\ \cdots (A) \ \dfrac {\partial p}{\partial y}= \dfrac {-y - (x-y)\times 1}{xy^2}= \dfrac {-x}{xy^2}- - \dfrac {1}{y^2}\ \cdots (B) \ \dfrac {\partial p}{\partial z}=0 \ \cdots (C) \ \dfrac {\partial q}{\partial x}= \dfrac {-1(x)- (z-x)\times 1}{x^2z} = \dfrac {-z}{x^2z}= - \dfrac {1}{x^2} \ \cdots (D) \ \dfrac {\partial q}{\partial y}=0 \ \cdots (E) \ \dfrac {\partial q}{\partial z}= \dfrac {1(z)-(z-x)\times 1}{xz^2}= \dfrac {x}{xz^2}= \dfrac {1}{z^2}\ \cdots (F)$ u is the function of p,q and p,q are the functions of x,y,z $\therefore \dfrac {\partial u}{\partial x}= \dfrac {\partial u}{\partial p}\dfrac {\partial p}{\partial x }+ \dfrac {\partial u}{\partial q} \dfrac {\partial q}{\partial x}$ $\dfrac {\partial u}{\partial x}= \dfrac {\partial u}{\partial p}\times \dfrac {1}{x^2}+ \dfrac {\partial u}{\partial q} \left ( - \dfrac {1} {x^2}\right ) \ \ \ {from \ (A) \ and \ (E)}$ $\therefore x^2 \dfrac {\partial u}{\partial x}= \dfrac {\partial u}{\partial p} - \dfrac {\partial u}{\partial q} \ \cdots (G) $ $\dfrac {\partial u}{\partial y}= \dfrac {\partial u}{\partial x}= \dfrac {\partial u}{\partial p}\dfrac {\partial p}{\partial y}+ \dfrac {\partial u}{\partial q}\dfrac {\partial q}{\partial y} $ $\dfrac {\partial u}{\partial y}= \dfrac {\partial u}{\partial p}\left ( - \dfrac {1}{y^2} \right )+ \dfrac {\partial u}{\partial q} (0) \ \ \ {from \ (B) \ and \ (E)} $ $\therefore y^2 \dfrac {\partial u}{\partial y}= - \dfrac {\partial u}{\partial p} \ \cdots (H)$ $\dfrac {\partial u}{\partial z}= \dfrac {\partial u}{\partial p}\dfrac {\partial p}{\partial z}+ \dfrac {\partial u}{\partial q} \dfrac {\partial q}{\partial z} $ $\dfrac {\partial u}{\partial z}= \dfrac {\partial u}{\partial p}(0)+ \dfrac {\partial u}{\partial q} \left ( \dfrac {1}{z^2} \right ) \ \ \ \ { from \ (C) \ and \ (F)}$ $\therefore z^2 \dfrac {\partial u}{\partial z} = \dfrac {\partial u}{\partial q} \ \cdots (I) $ $Adding \ (G),(H)\ and \ (I)$ $x^2 \dfrac {\partial u}{\partial x}+ y^2 \dfrac {\partial u}{\partial y}+ z^2 \dfrac {\partial u}{\partial z} = \dfrac {\partial u}{\partial p}- \dfrac {\partial u}{\partial q}- \dfrac {\partial u}{\partial p}+ \dfrac {\partial u}{\partial q} \ \therefore x^2 \dfrac {\partial u}{\partial x} + y^2 \dfrac {\partial u}{\partial y}+ z^2 \dfrac {\partial u}{\partial z}=0 $