$dxdy=rdrd\theta$ I'm trying to show that $dx\,dy=r\,dr\,d\theta$ using differentials.
$x=r\cos(\theta)$ and $y=r\sin(\theta)$
thus $dx=\cos(\theta)dr-r\sin(\theta)d\theta$ and $dy=\sin(\theta)dr+r\cos(\theta)d\theta$
$\begin{align}dx\,dy&=(\cos(\theta)dr-r\sin(\theta)d\theta)(\sin(\theta)dr+r\cos(\theta)d\theta)\\&
=\cos(\theta)\sin(\theta)
dr^2+r\cos^2(\theta)drd\theta-r\sin^2(\theta)d\theta dr-r^2\cos(\theta)\sin(\theta)d\theta^2\\&=\cos(\theta)\sin(\theta)
dr^2-r^2\cos(\theta)\sin(\theta)d\theta^2+rdrd\theta(1-\sin(\theta)^2-\sin(\theta)^2)\end{align}$
If my calculations are correct, $\cos(\theta)\sin(\theta)
dr^2-r^2\cos(\theta)\sin(\theta)d\theta^2-2 \sin(\theta)^2rdrd\theta=0$ but how am I suppose to show that?

You are missing the important point on the difference between the partition elements in Cartesian and Polar systems. 
While $dxdy$ is the area of a rectangle $rdrd\theta $ is the area of the curved section between circles of radii $r$ and $r+dr$ and the central angle of $d\theta$ 
The so called Jacobian gives you the multiplier of your transformation. The Jacobian is the determinant of a matrix whose terms are partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$ 
