Answer
$dL=ze^{-y^2-z^2}dx-2xyze^{-y^2-z^2}dy+e^{-y^2-z^2}(x-2xz^2)dz$
Work Step by Step
Given: $L=xze^{-y^2-z^2}$
Write the differential form such as follows:
$dL=\dfrac{\partial L}{\partial x} \times dx +\dfrac{\partial L}{\partial y} \times dy+\dfrac{\partial L}{\partial z} dz$
Write the partial derivatives w.r.t $x$ ; $y$ and $z$ as follows:
$dL=\dfrac{\partial L}{\partial x} dx +\dfrac{\partial L}{\partial y} dy+\dfrac{\partial L}{\partial z} dz\\=ze^{-y^2-z^2}dx+(-2xyze^{-y^2-z^2})dy+e^{-y^2-z^2}(x-2xz^2)dz\\=ze^{-y^2-z^2}dx-2xyze^{(-y^2-z^2)}dy+e^{(-y^2-z^2)}(x-2xz^2)dz$
hence, we have $dL=ze^{-y^2-z^2}dx-2xyze^{-y^2-z^2}dy+e^{-y^2-z^2}(x-2xz^2)dz$
