Prove that a local diffemorphism $f$ preserves the Gauss measure if and only if some condition is met I was reading through these lecture notes and found this exercise (page 6)

Let $f:U\rightarrow U$ be a local $C^1$ diffeomorphism, and let $\rho$ be a continuous function. Show that $f$ preserves the measure $\mu=\rho m$ if and only if
  $$\sum_{x\in f^{-1}(y)} \frac{\rho(x)}{|\mbox{det}Df(x)|}=\rho(y).$$

From the context, it looks like $m$ is the Lebesgue measure and $\mu$ the Gauss measure. How does one go about proving something like this?
EDIT: for context, the Gauss measure is defined as follows
$$\mu(B) = \frac{1}{\log 2}\int_B \frac{1}{1+x}dx.$$
For a function $f$ to preserve a measure $\mu$ means that
$$\mu(f^{-1}(B))=\mu(B)$$
for any measurable $B$.
The lecture notes don't say this explicitly, but I'd imagine $U$ is some open set.

I see the confusion: the statement you are trying to prove is general and unrelated to the Gauss measure.
Apply the area formula. For any integrable function $u : U \to \mathbb R$ you have $$\int_{f^{-1}(U)} u(x) |\det Df(x)| \, dx = \int_U \sum_{x \in f^{-1}(y)} u(x) \, dy.$$
If $B \subset U$ is measurable you can take $\displaystyle u(x) = \frac{\chi_{f^{-1}(B)}(x)}{|\det Df(x)|} \rho(x)$ so that the left hand side reduces to $\displaystyle \int_{f^{-1}(B)} \rho(x) \, dx = \mu(f^{-1}(B))$.
Now focus on the right side. $\chi_{f^{-1}(B)}(x) = 1$ if and only if $f(x) \in B$ so that the integral equals $$\int_B \sum_{x \in f^{-1}(y)} \frac{\rho (x)}{|\det Df(x)|} \, dx.$$ This equals $\mu(B)$ if and only if the integrand equals $\rho(y)$.
