Why $\frac{d\theta}{dt}=\frac{x}{r^{2}}\frac{dy}{dt}-\frac{y}{r^{2}}\frac{dx}{dt}$?

As the title shows this question concerns nothing but chain rule. We now have:

$$\frac{d\theta}{dt}=\frac{x}{r^{2}}\frac{dy}{dt}-\frac{y}{r^{2}}\frac{dx}{dt}$$

I am assuming by chain rule we have $$\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$$

But we have $$\theta=\arccos[\frac{x}{r}]=\arcsin[\frac{y}{r}]$$

Thus taking the derivative we should assume $$\frac{d\theta}{dx}=-\frac{1}{y},\frac{d\theta}{dy}=\frac{1}{x}$$ because $$\frac{d}{dx}\arccos[\frac{x}{r}]=-\frac{1}{r\sqrt{1-\frac{x^{2}}{r^{2}}}}=-\frac{1}{\sqrt{r^{2}-x^{2}}}=-\frac{1}{y}$$

However we know $$-\frac{y}{r^{2}}\not=-\frac{1}{y}$$ I computed this a few times but do not know where I got wrong. The relationship in the title is in Berkeley Problems in Mathematics.

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Try $\theta=\arctan(y/x)$. – Gerry Myerson Jul 21 '12 at 23:46
@GerryMyerson: Thanks. I still do not know where I had the mistake, but obviously you are right... – Bombyx mori Jul 21 '12 at 23:51
In the second equality, by the chain rule you have either ·$$\frac{{d\theta }}{{dt}} = \frac{{d\theta }}{{dx}}\frac{{dx}}{{dt}}$$ or $$\frac{{d\theta }}{{dt}} = \frac{{d\theta }}{{dy}}\frac{{dy}}{{dt}}$$ – Pedro Tamaroff Jul 21 '12 at 23:57
I see. Since I am assuming $\theta=f(r,x)$ with $r$ constant there is no $y$ involved. Thanks. – Bombyx mori Jul 22 '12 at 0:01
I don't think you computed $\frac{d\theta}{dx}$ correctly. – Ink Jul 22 '12 at 0:04

After the clarification (polar coordinates) and summarizing the answers given in the comments above: $$x=r\cos\theta\,\,,\,\,y=r\sin\theta\,\,\,,\,\,r\geq0\,,\,\,\theta\in [0,2\pi]$$ I'm assuming the radius is always non-negative, though not all do this.
From the above, $\,\theta=\arctan\frac{y}{x}\,\,,\,x\neq 0$ (the case $\,x=0\,$ is an easy particular case depending on the sign of $\,y\,$), so if both rectangular coordinates are derivable functions of some parameter $\,t\,$, we'd get: $$\frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}+\frac{d\theta}{dy}\frac{dy}{dt}$$ $$\frac{d\theta}{dx}=-\frac{y}{x^2}\frac{1}{1+\left(\frac{y}{x}\right)^2}=-\frac{y}{x^2+y^2}$$ $$\frac{d\theta}{dy}=\frac{1}{x}\frac{1}{1+\left(\frac{y}{x}\right)^2}=\frac{x}{x^2+y^2}$$
Observe that writing the expressions for $\,x,y\,$ from the beginning you get two differential equations. I'll leave this here as I'm not completely sure whether this already answers your question.