Answer
$(2\pi) \cos (2 \pi x+y-3z), \cos (2 \pi x+y-3z), -3\cos (2 \pi x+y-3z)$
Work Step by Step
Given: $h=\sin (2 \pi x+y-3z)$
Now, $\dfrac{\partial h}{\partial x}=\dfrac{\partial }{\partial x} [\sin (2 \pi x+y-3z)]=(2\pi) \cos (2 \pi x+y-3z)$
and
$\dfrac{\partial h}{\partial y}=\dfrac{\partial }{\partial y} [\sin (2 \pi x+y-3z)]=(1) \cos (2 \pi x+y-3z)=\cos (2 \pi x+y-3z)$
and
$\dfrac{\partial h}{\partial z}=\dfrac{\partial }{\partial z} [\sin (2 \pi x+y-3z)]=(-3) \cos (2 \pi x+y-3z)=-3\cos (2 \pi x+y-3z)$
Hence, $(2\pi) \cos (2 \pi x+y-3z), \cos (2 \pi x+y-3z), -3\cos (2 \pi x+y-3z)$
