Partial Differentiation Q

Ok here goes....

$\displaystyle

z=x^{2}+y^{2} $

$\displaystyle x=rcos\vartheta$

$\displaystyle y=r sin\vartheta$

Find:

$\displaystyle \frac{\partial z}{\partial x}_{y},

\frac{\partial z}{\partial \vartheta}_{x},

\frac{\partial z}{\partial r}_{y},

\frac{\partial z}{\partial r}_{ \vartheta},

$

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$\displaystyle \frac{\partial z}{\partial x}_{y} = 2x$

this seems right to me (though i'm not sure if i'm supposed to use a chain rule), it's the next ones i'm not sure about....

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$\displaystyle

\frac{\partial z}{\partial \vartheta}_{x}= $

I've been at this one for hours, i think i can use the following, but i'm getting nowhere.

$\displaystyle dz= \frac{\partial z}{\partial x}_{y}dx + \frac{\partial z}{\partial y}_{x}dy$

$\displaystyle dx= \frac{\partial x}{\partial r}_{\vartheta}dr + \frac{\partial x}{\partial \vartheta}_{r}d\vartheta$

$\displaystyle dy= \frac{\partial y}{\partial r}_{\vartheta}dr + \frac{\partial y}{\partial \vartheta}_{r}d\vartheta$

so....

$\displaystyle dz= \frac{\partial z}{\partial x}_{y} \left[ \frac{\partial x}{\partial r}_{\vartheta}dr + \frac{\partial x}{\partial \vartheta}_{r}d\vartheta\right]+ \frac{\partial z}{\partial y}_{x}\left[ \frac{\partial y}{\partial r}_{\vartheta}dr + \frac{\partial y}{\partial \vartheta}_{r}d\vartheta\right]$

can i divide through by $\displaystyle \partial \vartheta_{x}$ and then work it all out to get

$\displaystyle

\frac{\partial z}{\partial \vartheta}_{x} $ ?

please help, i don't have a clue what i'm doing!!! :confused: