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