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.