Answer
$$\frac{\partial w}{\partial s}=2s\cos2t,\frac{\partial w}{\partial t}=-2s^2\sin2t$$
For $s=3$ and $t=\pi/4$:
$$\frac{\partial w}{\partial s}=0,\frac{\partial w}{\partial t}=-18$$
Work Step by Step
To find both partial derivatives we will use Chain Rule.
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial }{\partial x}(x^2-y^2)\frac{\partial }{\partial s}(s\cos t)+\frac{\partial }{\partial y}(x^2-y^2)\frac{\partial }{\partial s}(s\sin t)=2x\cos t-2y\sin t$$
Expressing this in terms of $s$ and $t$ we have:
$$\frac{\partial w}{\partial s}=2x\cos t-2y\sin t=2s\cos t\cdot \cos t-2s\sin t\cdot \sin t=2s(\cos^2t-\sin^2t)=2s\cos2t$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial }{\partial x}(x^2-y^2)\frac{\partial }{\partial t}(s\cos t)+\frac{\partial }{\partial y}(x^2-y^2)\frac{\partial }{\partial t}(s\sin t)=2x(-s\sin t)-2ys\cos t=-2sx\sin t-2sy\cos t$$
Expressing this in terms of $s$ and $t$ we have:
$$\frac{\partial w}{\partial t}=-2sx\sin t-2sy\cos t=-2s\cdot s\cos t\cdot\sin t-2s\cdot s\sin t\cdot \cos t=-4s^2\cos t\sin t=-2s^2\sin 2t$$
Now we will evaluate these partial derivatives at the given values $s=3$ and $t=\pi/4$:
$$\frac{\partial w}{\partial s}=2s\cos 2t=2\cdot3\cos2\frac{\pi}{4}=6\cos\frac{\pi}{2}6\cdot0=0$$
$$\frac{\partial w}{\partial t}=-2s^2\sin2t=-2\cdot3^2\sin2\frac{\pi}{4}=-2\cdot9\sin\frac{\pi}{2}=-18\cdot1=-18$$
