Answer
\begin{aligned} d w =-e^{y} \sin x \ d x+e^{y} \cos x \ d y+2 z \ d z \end{aligned}
Work Step by Step
Given $$ w =e^{y} \cos x+z^{2}$$
Since $$dw=\frac{\partial w}{\partial x} dx+\frac{\partial w}{\partial y} dy+\frac{\partial w}{\partial z} dz,$$
$$\frac{\partial w}{\partial x} =-e^{y} \sin x ,$$
$$\frac{\partial w}{\partial y} =e^{y} \cos x$$
and
$$\frac{\partial w}{\partial z} = 2z $$
then we get
\begin{aligned} d w =-e^{y} \sin x \ d x+e^{y} \cos x \ d y+2 z \ d z \end{aligned}
