Answer
$$\frac{\partial w}{\partial s}=5\cos(5s-t),\frac{\partial w}{\partial t}=-cos(5s-t)$$
For $s=0$ and $t=\pi/2$:
$$\frac{\partial w}{\partial s}=0,\frac{\partial w}{\partial t}=0$$
Work Step by Step
We will find both partial derivatives using the Chain Rule.
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial}{\partial x}(\sin(2x+3y))\frac{\partial}{\partial s}(s+t)+\frac{\partial}{\partial y}(\sin(2x+3y))\frac{\partial}{\partial s}(s-t)=\cos(2x+3y)\frac{\partial}{\partial x}(2x+3y)\cdot1+\cos(2x+3y)\frac{\partial}{\partial y}(2x+3y)\cdot1=\cos(2x+3y)\cdot2+\cos(2x+3y)\cdot3=5\cos(2x+3y)$$
Expressing this in terms of $s$ and $t$ we get:
$$\frac{\partial w}{\partial s}=5\cos(2x+3y)=5\cos(2(s+t)+3(s-t))=5\cos(2s+2t+3s-3t)=5\cos(5s-t)$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial }{\partial x}(\sin(2x+3y))\frac{\partial }{\partial t}(s+t)+\frac{\partial }{\partial y}(\sin(2x+3y))\frac{\partial }{\partial t}(s-t)=
\cos(2x+3y)\frac{\partial }{\partial x}(2x+3y)\cdot1+\cos(2x+3y)\frac{\partial }{\partial y}(2x+3y)\cdot(-1)=\cos(2x+3y)\cdot2-\cos(2x+3y)\cdot3=-\cos(2x+3y)$$
Expressing this in terms of $s$ and $t$ we have:
$$\frac{\partial w}{\partial t}=-\cos(2x+3y)=-\cos(2(s+t)+3(s-t))=-\cos(5s-t)$$
Now we will evaluate these partial derivatives at the given values $s=0$ and $t=\pi/2$:
$$\frac{\partial w}{\partial s}=5\cos(5s-t)=5\cos(5\cdot0-\frac{\pi}{2})=5\cos(-\frac{\pi}{2})=5\cdot0=0$$
$$\frac{\partial w}{\partial t}=-\cos(5s-t)=-\cos(5\cdot0-\frac{\pi}{2})=-\cos(-\frac{\pi}{2})=0$$
