Showing that $\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)$ I'm trying to show that

$$
\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)
$$

but am having trouble figuring out exactly how to approach the problem. I've tried a few tricks with switching to polar coordinates or trying to make use of right triangles, but have had no luck. Could anyone point me in the right direction here?
Hint: 
$$ \frac{d}{dt}\tan^{-1}(t) = \frac{1}{1+t^2} $$
$$ d f(x,y) = \frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy $$
Using the Hint from above:
\begin{align}
d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) &amp;= \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial x} dx + \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial y} dy \\
&amp;= \frac{1}{1+\frac{x^2}{y^2}}\frac{1}{y}dx + \frac{1}{1+\frac{x^2}{y^2}}\left(-\frac{x}{y^2} \right)dy\\
&amp;=\frac{y}{x^2+y^2}dx-\frac{x}{y^2+x^2}dy
\end{align}