Im a bit confused with the push forward formula https://en.wikipedia.org/wiki/Pushforward_measure
Let $\rho$ be a probability density associated to a probability measure $\mu$ on $\mathbb{R}^d$, i.e "$\rho(x)dx=d\mu(x)$". Let $f:\mathbb{R}^d\to\mathbb{R}^d$ and denote $f_{\#}\mu$ be the push forward of $\mu$ by $f$, and $f_{\#}\rho$ the associated density pushed forward.
Then $\textbf{is it true that}$ for any integral
$$ \int_{\mathbb{R}^d} \big( g \circ f_{\#}\rho \big) (x) df_{\#}\mu(x) =\int_{\mathbb{R}^d} \big( g\circ \rho \big) \rho(x)dx $$
What does $f_\sharp\rho$ mean? Is it the density of $f_\sharp\mu$ with respect to the Lebesgue measure? In the general case there does not necessarily exist such a density, e.g. $f:x\mapsto0$. Even if $f_\sharp\rho$ exists, your equation has no reason to hold. Take for instance $d=1$, $\mu$ equal to the Lebesgue measure on $(0,1)$ and $f:x\mapsto2x$, so that $\rho(x)=1_{(0,1)}(x)$, $f_\sharp\mu$ is half of the Lebesgue measure on $(0,2)$ and $f_\sharp\rho(x)=\frac12\times1_{(0,2)}(x)$. Let $g:\mathbb R\to\mathbb R$ be any nonnegative measurable map such that $g(x)=0$ on $[1,2]$ and $g(x)>0$ on $(0,1)$. Then in your equation, the left-hand side is positive whereas the right-hand side is zero.