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Consider the transformation $F$ of $mathbb R^2setminus{(0,0)}$ onto itself defined as
$$
F(x, y):=left( frac{x}{x^2+y^2}, frac{y}{x^2+y^2}right).$$
Its Jacobian matrix is
$$tag{1}
begin{bmatrix} frac{y^2-x^2}{(x^2+y^2)^2} & -frac{2xy}{(x^2+y^2)^2} \ -frac{2xy}{(x^2+y^2)^2} & frac{x^2-y^2}{(x^2+y^2)^2} end{bmatrix},quad text{and its determinant equals} frac{-1}{(x^2+y^2)^2}.$$
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 $$tag{2}tilde{F}:=phi^{-1}circ Fcirc phi;$$ then, by an easy direct computation, $$tilde{F}(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tilde{F}$ is, thus, $$tag{!}begin{bmatrix} frac{-1}{r^2} & 0 \ 0 & 1end{bmatrix} , quad text{and its determinant equals } frac{-1}{r^2}.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tilde{F}$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag{!!} frac{-1}{r^2}=frac{-1}{x^2+y^2}.$$

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
$$
begin{bmatrix} frac{x}{sqrt{x^2+y^2}} & frac{y}{sqrt{x^2+y^2}} \ -frac{y}{x^2+y^2} & frac{x}{x^2+y^2}end{bmatrix} begin{bmatrix} frac{y^2-x^2}{(x^2+y^2)^2} & -frac{2xy}{(x^2+y^2)^2} \ -frac{2xy}{(x^2+y^2)^2} & frac{x^2-y^2}{(x^2+y^2)^2} end{bmatrix}begin{bmatrix} costheta & -rsintheta \ sin theta & rcos thetaend{bmatrix} = begin{bmatrix} -frac1{r^2} & 0 \ 0 & frac{1}{r^2}end{bmatrix},$$
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, $D(phi(frac{1}{r}, costheta)) × Dtilde{F}(r, theta)= DF times D(phi(r, theta))$

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
$$ F^star(omega_{rm Cart}) = - frac{1}{(x^2+y^2)^2}omega_{rm Cart}.$$

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:
begin{align}
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^2}omega_{rm Polar} right) \ &= - frac{1}{r^4} left(romega_{rm Polar} right) \ &= - frac 1 {r^4} omega_{rm Cart}
end{align}
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

$$ D(phi^{-1})|_{Fcirc phi(r, theta)} = begin{bmatrix} frac{frac{x}{x^2 + y^2}}{sqrt{left(frac{x}{x^2 + y^2} right)^2+left( frac{y}{x^2 + y^2}right)^2}} & frac{frac{y}{x^2 + y^2}}{sqrt{left(frac{x}{x^2 + y^2} right)^2+left( frac{y}{x^2 + y^2}right)^2}} \ -frac{frac{y}{x^2 + y^2}}{left(frac{x}{x^2 + y^2} right)^2+left( frac{y}{x^2 + y^2}right)^2} & frac{frac{x}{x^2 + y^2}}{left(frac{x}{x^2 + y^2} right)^2+left( frac{y}{x^2 + y^2}right)^2}end{bmatrix} = begin{bmatrix}costheta & sin theta \ - rsin theta & rcostheta end{bmatrix}$$
which is not the inverse of $(Dphi)|_{(r, theta)}$.