$\Large{ We \; have \\ f_1 : u-xyz = 0 , \\ f_2 : v-x^2-y^2-z^2=0, \\ f_3: w-x-y-z=0 \\ \therefore {\delta x \over \delta u }= - \bigg[ { \delta (f_1,f_2,f_3) \over \delta(u,y,z)} \bigg/ \; {\delta(f_1,f_2,f_3) \over \delta (x,y,z)} \bigg] \\ Now, \; \; { \delta (f_1,f_2,f_3) \over \delta(u,y,z)} \; = \;\; \begin{vmatrix} 1 & -xz & \; -xy \\ 0 & -2y & -2z \\ 0 & -1 & -1 \end{vmatrix} \; \; = 2(y-z) \\ And \; \; \; { \delta (f_1,f_2,f_3) \over \delta(x,y,z)} \; = \;\; \begin{vmatrix} -yz & -xz & -xy \\ -2x & -2y & -2z \\ -1 & -1 & -1 \end{vmatrix} \\ = -2 \; \begin{vmatrix} yz & zx & xy \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} \\ = -2 \begin{vmatrix} yz & z(x-y) & x(y-z) \\ x & y-x & z-y \\ 1 & 0 & 0 \end{vmatrix} \\ By \; C_2 - C_1 \; and \; C_3 - C_2 \\ = -2 (x-y)(y-z) \; \;\begin{vmatrix} yz & z & x \\ x & -1 & -1 \\ 1 & 0 & 0 \end{vmatrix} \\ = 2 (x-y) (y-z)(z-x) \\ \therefore {\delta x \over \delta u} = {2(y-z) \over 2(x-y)(y-z)(z-x)} \\ \therefore {\delta x \over \delta u} = {1 \over (x-y)(z-x)} \\ }$

Hence proved.