Answer
$$\frac{\partial w}{\partial r} = 2\pi$$ $$\frac{\partial w}{\partial \theta} =-2\pi$$
Work Step by Step
1. According to the chain rule:
$$\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial r}$$ $$\frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial \theta} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial \theta}$$
$$\frac{\partial w}{\partial x} = \frac{\partial (xy + yz + zx)}{\partial x}= y + z$$ $$\frac{\partial w}{\partial y} = \frac{\partial (xy + yz + zx)}{\partial y}= x + z$$ $$\frac{\partial w}{\partial z} = \frac{\partial (xy + yz + zx)}{\partial z}= y + x$$ $$\frac{\partial x}{\partial r} = \frac{\partial (rcos \space \theta)}{\partial r} = cos \space \theta$$ $$\frac{\partial x}{\partial \theta} = \frac{\partial (rcos \space \theta)}{\partial \theta} = -rsin \space \theta$$
$$\frac{\partial y}{\partial r} = \frac{\partial (rsin \space \theta)}{\partial r} = sin \space \theta$$ $$\frac{\partial y}{\partial \theta} = \frac{\partial (rsin \space \theta)}{\partial \theta} = rcos \space \theta$$
$$\frac{\partial z}{\partial r} = \frac{\partial (r \theta)}{\partial r} = \theta$$ $$\frac{\partial z}{\partial \theta} = \frac{\partial (r \theta)}{\partial \theta} = r$$
Therefore:
$$\frac{\partial w}{\partial r} =(y + z)(cos \space \theta) + (x + z)(sin \space \theta) + (y + x)(\theta)$$ $$\frac{\partial w}{\partial \theta} = (y+z)(-rsin \space \theta) + (x+z)(rcos \space \theta) + (y + x)(r)$$
2. Calculate the value for each partial derivate.
$r = 2$
$\theta = \pi/2$
$x = rcos(\theta) = (2)(cos \space (\pi/2)) = 0$
$y =rsin(\theta) = (2)(sin(\pi/2)) = 2$
$z = r \theta = (2)(\pi/2) = \pi$
$$\frac{\partial w}{\partial r} =(2 + \pi)(cos \space (\pi/2)) + (0 + \pi)(sin \space (\pi /2)) + (2 + 0)(\pi/2)$$ $$\frac{\partial w}{\partial r} = \pi + \pi = 2\pi$$ $$\frac{\partial w}{\partial \theta} =(2 + \pi)(-2 sin \space (\pi/2)) + (0 + \pi)(2cos \space (\pi /2)) + (2 + 0)(2)$$ $$\frac{\partial w}{\partial \theta} = -4 - 2\pi +4 = -2\pi$$
