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Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$
Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.

Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$

The result (!!) is off by a factor of $r^-2$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.

Can you help me spot the mistake?

The Jacobians of the two functions aren't equal by the chain rule.

In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$

I don't think there is any contradiction here.

Consider the volume form
$$ omega_rm Cart = dx wedge dy.$$
Your first calculation shows that the pullback $F^star(omega_rm Cart)$ is given by
$$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$

Now consider the volume form
$$ omega_rm Polar = dr wedge dtheta.$$
Your second calculation shows that

$$ F^star(omega_rm Polar)=-frac 1 r^2 omega_rm Polar. $$

We can use this to recompute $F^star(omega_rm Cart)$. In view of the fact that
$$ omega_rm Cart = r omega_rm Polar,$$
we have:
beginalign
F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
endalign
which is consistent with the first calculation!

As for the application of the chain rule, we have:
$$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$

The key point is that you must evaluate $D(phi^-1)$ at the point $left(frac x (x^2 +y^2), frac y (x^2 + y^2)right)$, not at the point $(x, y)$.

This is equal to

$$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
which is not the inverse of $(Dphi)|_(r, theta)$.