Answer
$1+2uvw-u^2-v^2-w^2$
Work Step by Step
Since, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}$
Given: $x=u+vw$
Here, we have $\dfrac{\partial x}{\partial u}=1; \dfrac{\partial x}{\partial v}=w$ and $ \dfrac{\partial x}{\partial w}=v$
Given that $y=v+wu$
Thus, $\dfrac{\partial y}{\partial u}=w; \dfrac{\partial y}{\partial v}=1$ and $ \dfrac{\partial y}{\partial w}=u$
Given: $z=w+uv$
Thus, $\dfrac{\partial z}{\partial u}=v; \dfrac{\partial z}{\partial v}=u$ and $ \dfrac{\partial z}{\partial w}=1$
Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}=\begin{vmatrix} 1&w&v\\w&1&u\\v&u&1\end{vmatrix}=1(1-u^2)-w(w-uv)+v(uw-v)=1+2uvw-u^2-v^2-w^2$
