Let $f_1=U-xyz----------(1) \\ f_2 = V-x^2-y^2-z^2------(2) \\ f_3=w-x-y-z-------(3)$

Differentiating $f_1,f_2,f_3$ w.r.t X we get

$∴ \dfrac {∂f_1}{∂x} = f_1 x=0-yz=-yz \\ \dfrac {∂f_2}{∂x} =f_2 x=0-2x-0-0=-2x \\ \dfrac {∂f_3}{∂x} = f_3 x=0-1-0=-1 $

Similarly

Differentiating w.r.t Y

$f_1 Y=-xz \\ f_2 Y=-2y \\ f_3 Y=-1 $

Differentiating w.r.t Z

$f_(1 ) z=-xy \\ f_(2 ) z=-2z \\ f_(3 ) z=-1$

Differentiating w.r.t U

$f_1 U=1-0=1 \\ f_2 U=0 \\ f_3 U=0 \\ \therefore \dfrac {d(f_1,f_2,f_3)}{d(U,Y,Z)} = \begin{vmatrix} f_1u & f_1y& f_1z\\ f_2u& f_2y & f_2z \\ f_3u & f_3y& f_3z \\ \end {vmatrix} \\ =\begin{vmatrix} 1&-xz & -xy \\ 0 & -2y & -2z\\ 0 &-1 &-1 \\ \end {vmatrix} \\ =2y-2z \\ =2(y-z) \\ Also \space \space \dfrac {d(f_1,f_2,f_3)}{d(U,Y,Z)} = \begin{vmatrix} f_1x & f_1y& f_1z\\ f_2x& f_2y & f_2z \\ f_3x & f_3y& f_3z \\ \end {vmatrix} \\= \begin{vmatrix} -yz & -xz& -xy\\ -2x& -2y & -2z \\ -1 & -1& -1 \\ \end {vmatrix} \\ =-yz(+2y-2z)+xz(2x-2z)-xy(2x-2y) \\ =-2y^2 z+2yz^2+2x^2 z-2xz^2-2x^2 y+2xy^2$

Adjusting such that to make combination of y & z

$=2[x^2 z-x^2 y-xz^2+xy^2+yz^2-y^2 z] \\ =2[x^2 (z-y)-x(z^2-y^2 )+yz(z-y)] \\ =2(z-y)[x^2-x(z+y)+yz] \\ =2[z-y][x^2-xz-xy+yz] \\ =2[z-y][x(x-z)-y(x-z)] \\ =2(z-y)(x-z)(x-y) \\ ∴ \dfrac {∂x}{∂u} = [\dfrac {\dfrac {∂(f_1,f_2,f_3 )}{∂(u,y,z)}}{\dfrac {∂(f_1,f_2,f_3 )}{∂(x,y,z)}}] × (-1) \\ \dfrac {∂x}{∂u} = -[\dfrac {2(y-z)}{2(z-y)(x-z)(x-y)} ] \\ \dfrac {∂x}{∂u} = + \dfrac 1{(x-y)(x-z) }$

Hence Proved.