u = $ 1-x \\ \; \\ Diff. \; partially \; w.r.t. \; x, \\ \; \\ \therefore \frac{\partial u }{\partial x} \; = \; -1 \\ \; \\ Similarly \; \frac{\partial u }{\partial y} \; = \; 0 \; \; and \; \; \frac{\partial u }{\partial z} \; = \; 0 \\ \; \\ \; \\ v=x(1-y) \\ \; \\ Diff. \; partially \; w.r.t. \; x, \\ \; \\ \therefore \frac{\partial v }{\partial x} \; = \; 1-y \\ \; \\ \frac{\partial v }{\partial y} \; = \; 0-x \; = \; -x \\ \; \\ \frac{\partial v }{\partial z} \; = \; 0 \\ \; \\ \; \\ w = xy(1-z) \\ \; \\ Diff. \; partially \; w.r.t. \; x, \\ \; \\ \therefore \frac{\partial w }{\partial x} \; = \; y-yz \\ \; \\ Similarly \; \frac{\partial w }{\partial y} \; = \; x-xz \\ \; \\ \frac{\partial w }{\partial z} \; = \; -xy \\ \; \\ \; \\ Jacobian|J| \; = \; \frac{\partial (u,v,w)}{\partial (x,y,z)} \;= \; \left| \begin{array}{ccc} \dfrac{\partial u }{\partial x} & \dfrac{\partial u }{\partial y} & \dfrac{\partial u }{\partial z} \\ \; \\ \dfrac{\partial v }{\partial x} & \dfrac{\partial v }{\partial y} & \dfrac{\partial v }{\partial z} \\ \; \\ \dfrac{\partial w }{\partial x} & \dfrac{\partial w }{\partial y} & \dfrac{\partial w }{\partial z} \end{array} \right| \;= \; \left| \begin{array}{ccc} -1 & 0 & 0 \\ \; \\ 1-y & -x & 0 \\ \; \\ y-yz & x-xz & -xy \end{array} \right| \\ \; \\ \; \\ \dfrac{\partial (u,v,w)}{\partial (x,y,z)} \;= \; -1[(-x)(-xy)-0(x-xz)] \; - \; 0[(1-y)(-xy)-0(y-yz)] \; + \; \\ [(1-y)(x-xz)-(-x)(y-yz)] \\ \; \\ = \; -1(x^2y-0)-0+0 \\ \; \\ - \; -x^2y \\ \; \\ \; \\ \frac{\partial (u,v,w) }{\partial (x,y,z)} \; = \; -x^2y \; \; \; \; \; Hence \; Proved. $