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
Given the function $L=xze^{-y^2-z^2}$
The differential form can be evaluated as follows:
$dL=\dfrac{\partial L}{\partial x} dx +\dfrac{\partial L}{\partial y} dy+\dfrac{\partial L}{\partial z} dz$
We need to find the partial derivatives w.r.t. $x$, $y$ and $z$ as follows:
$dL=ze^{-y^2-z^2}dx+(-2xyze^{-y^2-z^2})dy+e^{-y^2-z^2}(x-2xz^2)dz=ze^{-y^2-z^2}dx-2xyze^{-y^2-z^2}dy+e^{-y^2-z^2}(x-2xz^2)dz$
