In Aubin's book, $\textit{Some Nonlinear Problems in Riemannian Geometry}$, the following proof is given to show that $\mu$ (the Yamabe invariant) is conformally invariant.
The Equation (1) that is being referred to is:
I have encountered two difficulties with this proof.
First: When I evaluate $\nabla^i(\varphi\psi)\nabla_i(\varphi\psi)$, I get \begin{equation*} \nabla^i(\varphi\psi)\nabla_i(\varphi\psi) = \varphi^2 \nabla^i\psi\nabla_i\psi + \psi^2\nabla^i\varphi\nabla_i\varphi + \varphi\psi(\nabla^i\varphi\nabla_i\psi + \nabla^i\psi\nabla_i\varphi). \end{equation*} How do the last three terms in the above turn into $\varphi\psi^2\Delta\varphi$?
Second: After using equation (1), I get, \begin{equation*} J(\varphi\psi)=\frac{4\frac{n-1}{n-2}\big[\int_M \varphi^2 \nabla^i\psi\nabla_i\psi \, dV\big] + \int_M R' \psi^2 dV'}{\big[\int_M \psi^N dV'\big]^{2/N}}. \end{equation*} In the first integral on the numerator, I'm not sure how $\varphi^2$ is used to transform $dV$ into $dV'$.
Apologies if the answers to my difficulties are trivial (I'm still rather new to this type of material). Any help would be greatly appreciated. Thank you.
You're going to have to integrate by parts. Note also that Aubin uses the incorrect sign for $\Delta$, i.e. $\Delta u=-\mathrm{div}\,\nabla u$.
Integrating the second term by parts gives $$\int\psi^2|\nabla\varphi|^2=-\int\varphi\,\mathrm{div}(\psi^2\nabla\varphi)=\int\varphi\psi^2\Delta\psi-2\int\varphi\psi\nabla\psi\cdot\nabla\varphi.\tag{$*$}$$ Now notice that the last two terms in what you wrote are actually the same and equal together $2\varphi\psi\nabla\psi\cdot\nabla\varphi$. This cancels with the second term on the right in $(*)$.
As for the volume element, in local coordinates $dV=\sqrt{\det g}\,dx.$ Now we have $$\det g'=\det(\varphi^{4/(n-2)}g)=\varphi^{4n/(n-2)}\det g.$$ It follows that $dV'=\varphi^{2n/(n-2)}dV$. This looks wrong, but that's because you also have to account for the difference between $\nabla \psi\cdot\nabla\varphi$ and $\nabla'\psi\cdot\nabla'\varphi$. There's another factor of $\varphi$ that comes from this because of the conformal change in metric. If you work this out, you get $$\int \nabla'u\cdot\nabla'v\,dV'=\int\nabla u\cdot\nabla v\,\varphi^2\,dV$$ for any smooth functions $u$ and $v$. This should get you what you need.