Answer
$\dfrac{\partial w}{\partial x}=(\cos \theta) \dfrac{\partial w}{\partial r} -(\dfrac{ \sin \theta}{r}) \ \dfrac{\partial w}{\partial \theta}$
and
$\dfrac{\partial w}{\partial y}=(\sin \theta) \dfrac{\partial w}{\partial r} +(\dfrac{ \cos \theta}{r}) \ \dfrac{\partial w}{\partial \theta}$
Work Step by Step
To convert rectangular coordinates into polar coordinates, we will use:
$x =r \cos \theta$ and $y=r \sin \theta$ and $r^2=x^2+y^2$ and $\theta =\tan^{-1} (y/x)$
Now, we have: $\dfrac{\partial w}{\partial x}=\dfrac{\partial w}{\partial r} \ \dfrac{\partial r}{\partial x}+\dfrac{\partial w}{\partial \theta} \ \dfrac{\partial \theta}{\partial x} \\=\dfrac{\partial w}{\partial r} \ (\dfrac{x}{\sqrt {x^2+y^2}})+\dfrac{\partial w}{\partial \theta} \ (\dfrac{-y}{x^2+y^2}) \\=(\cos \theta) \dfrac{\partial w}{\partial r} -(\dfrac{ \sin \theta}{r}) \ \dfrac{\partial w}{\partial \theta}$
and
$\dfrac{\partial w}{\partial y}=\dfrac{\partial w}{\partial r} \ \dfrac{\partial r}{\partial y}+\dfrac{\partial w}{\partial \theta} \ \dfrac{\partial \theta}{\partial y} \\=\dfrac{\partial w}{\partial r} \ (\dfrac{y}{\sqrt {x^2+y^2}})+\dfrac{\partial w}{\partial \theta} \ (\dfrac{x}{x^2+y^2}) \\=(\sin \theta) \dfrac{\partial w}{\partial r} +(\dfrac{ \cos \theta}{r}) \ \dfrac{\partial w}{\partial \theta}$
