We always believe that $dxdy$ in cartesian coordinate is equivalent to $rdrd\phi$. So let's check:
$x=r\cos\phi$
$y=r\sin\phi$.
I differentiate the equations above to derive $dx$ and $dy$
$dx=dr\cos\phi-r\sin\phi d\phi$
$dy=dr\sin\phi+r\cos\phi d\phi$
then we have
$dxdy=dr^2\cos\phi \sin\phi+rdrd\phi \cos^2\phi-rdrd\phi \sin^2\phi-r^2d\phi^2\sin\phi \cos\phi$
which is not $rdrd\phi$. However, even I neglect the terms with $dr^2$ and $d\phi^2$, I have
$dxdy=rdrd\phi \cos^2\phi-rdrd\phi \sin^2\phi=rdrd\phi(\cos2\phi)$
again I did not reach $rdrd\phi$
Note for those who answered similar questions: Please do not refer again to the Jacobian. I am aware of such answers and I do not oppose them. However, my question is that why we see this inconsistency between different ways of deriving the differentiation.
An expansion on @AnginaSeng's comment:
If you compare a Taylor expansion $f(x+h)\in f(x)+hf^\prime(x)+\tfrac12h^2f^{\prime\prime}(x)+o(h^2))$ with the chain rule $f(x+dx)=f(x)+dx\cdot f^\prime(x)$, we get $dx^2=0$. If this is to work for all variables, $z=x+y$ gives$$0=dz^2=dx^2+dy^2+dxdy+dydx=dxdy+dydx.$$Really, this all should be written with $dx\wedge dy$ instead of $dxdy$; it's an antisymmetric product called a wedge product. Similarly, $dr^2=0$, $d\phi^2=0$ and $d\phi dr=-drd\phi$. Try it now.
More generally, you can use this idea to derive the Jacobian. For example, in $2$ dimensions a conversion between two coordinate systems looks like$$dx_1dx_2=(J_{11}dy_1+J_{12}dy_2)(J_{21}dy_1+J_{22}dy_2)=(J_{11}J_{22}-J_{12}J_{21})dy_1dy_2.$$