Answer
$2uvw$
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: $x=uv$
Here, we have $\dfrac{\partial x}{\partial u}=v; \dfrac{\partial x}{\partial v}=u$ and $ \dfrac{\partial x}{\partial w}=0$
Given: $y=vw$
Also, $\dfrac{\partial y}{\partial u}=0; \dfrac{\partial y}{\partial v}=w$ and $ \dfrac{\partial y}{\partial w}=v$
Given: $z=wu$
Also, $\dfrac{\partial z}{\partial u}=w; \dfrac{\partial z}{\partial v}=0$ and $ \dfrac{\partial z}{\partial w}=u$
Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}=\begin{vmatrix} v&u&0\\0&w&v\\w&0&u\end{vmatrix}$
or, $=v(wu-0)-u(0-vw) +0$
Hence, $Jacobian=2uvw$
