Answer
$$\frac{\partial H(x,y,z)}{\partial x}=\cos(x+2y+3z)$$
$$\frac{\partial H(x,y,z)}{\partial y}=2\cos(x+2y+3z)$$
$$\frac{\partial H(x,y,z)}{\partial z}=3\cos(x+2y+3z)$$
Work Step by Step
We will use chain rule to find these partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial H(x,y,z)}{\partial x}=\frac{\partial }{\partial x}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial}{\partial x}(x+2y+3z)=\cos(x+2y+3z)\cdot1=\cos(x+2y+3z)$$
Because $2y+3z$ is treated as a constant its' derivative is $0.$
The partial derivative with respect to $y$ is:
$$\frac{\partial H(x,y,z)}{\partial y}=\frac{\partial }{\partial x}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial}{\partial y}(x+2y+3z)=\cos(x+2y+3z)\cdot2=2\cos(x+2y+3z)$$
Here $x+3z$ is treated as a constant, so its' derivative is $0.$
The partial derivative with respect to $z$ is:
$$\frac{\partial H(x,y,z)}{\partial z}=\frac{\partial }{\partial z}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial }{\partial z}(x+2y+3z)=\cos(x+2y+3z)\cdot3=3\cos(x+2y+3z)$$
$x+2y$ is treated as a constant, so its' derivative is $0.$