Is every diffeomorphism conformally equivalent to a volume preserving diffeomorphism?

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Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Let $f:D \to D$ be a diffeomorphism.

Does there exist a smooth $h \in C^{\infty}(D)$ such that $h\cdot f$ is an area-preserving diffeomorphism of $D$?

Clearly, such an $h$ must send the entire boundary $\partial D$ either to $1$ or to $-1$. Another necessary condition is $\det\big(d (h\cdot f)\big) = 1$. Since

$$ d (h\cdot f)=h df+dh \otimes f=h df+f \cdot (\nabla h)^T, $$

by the matrix determinant lemma, at all points where $h \neq 0$, we have

$$ \det\big(d (h\cdot f)\big)=h^2\det(df) \big( 1+h^{-1}(\nabla h)^T ((df)^{-1} \cdot f)\big), $$

so we need to solve the following PDE for $h$

$$\det(df) \cdot \big( h^2+h(\nabla h)^T ((df)^{-1} \cdot f)\big)=1.$$