Answer
$ze^{-y^2-z^2}dx-2xyze^{-y^2-z^2}dy+e^{-y^2-z^2}(x-2xz^2)dz$
Work Step by Step
We are given that $L=xze^{-y^2-z^2}$
The differential form is written as:
$dL=(\dfrac{\partial L}{\partial x}) dx +(\dfrac{\partial L}{\partial y}) dy+(\dfrac{\partial L}{\partial z}) dz$
The partial derivatives with respect to $x$ and $y$ and $z$ are given as:
$dL=(\dfrac{\partial L}{\partial x}) dx +(\dfrac{\partial L}{\partial y}) dy+(\dfrac{\partial L}{\partial z}) dz=ze^{-y^2-z^2}dx+(-2xyze^{-y^2-z^2})dy+e^{-y^2-z^2}(x-2xz^2)dz$
or, $dL=ze^{-y^2-z^2}dx-2xyze^{-y^2-z^2}dy+e^{-y^2-z^2}(x-2xz^2)dz$
