I'm reading an article from $1988$ from Kuhnel, where I found the following:
$\textbf{Proposition}$. Let $(M, g)$ be a space of constant curvature $K$ and dimension $n \geq 3$ and $\psi: M \to \mathbb{R}$ a function, and $\bar{g} = \psi^{-2}g$.
$\textbf{Then the following are equivalent}$:
$(1)$ $(M, \bar{g})$ is a space of constant sectional curvature $\bar{K}$
$(2) $$(M, \bar{g})$ is Einstein
$(3)$ $\nabla^2 \psi = \cfrac{\Delta\psi}{n} \cdot g$
where $ \nabla^{2} \varphi=\nabla(\nabla \varphi) $ denotes the hessian of $\varphi$.
Here's the proof for $(3) \iff (1)$ (not verbatim):
Define $\varphi$ by $\psi \doteq e^{\varphi} $. Let $\sigma$ be the $2$-plane spanned by two vectors $X, Y$ which are orthonormal with respect to $g$. Let $\overline{K}_{\sigma}$ denote the sectional curvature of $\sigma$ in $(M, \bar{g})$. Then from $(1.2)$ we have:
$$ \begin{aligned} \psi^{-2} \overline{K}_{\sigma} &=\langle \overline{{R}}(X, Y) Y, X \rangle\\ &\color{red}{={K}+\nabla^{2} \varphi(Y, Y)+\nabla^{2} \varphi(X, X)+(Y \varphi)^{2}+(X \varphi)^{2}-\|\nabla \varphi\|^{2} }\\ &\color{red}{=K+\frac{2}{n} \cdot \frac{\Delta \psi}{\psi}-\|\nabla \varphi\|^{2} }\end{aligned} $$
and the result follows from Schur's theorem.
Now, $(1.2)$ is the following:
$$ \begin{aligned} \overline{R_{m}}(X, Y) Z &=R_{m}(X, Y) Z-\left[\langle X, Z\rangle H_{\varphi} Y-\langle Y, Z\rangle H_{\varphi} X\right] \\ &+\left[\nabla^{2} \varphi(Y, Z)+(Y \varphi)(Z \varphi)-\langle Y, Z\rangle\|\varphi\|^{2}\right] X \\ &-\left[\nabla^{2} \varphi(X, Z)+(X \varphi)(Z \varphi)-\langle X, Z\rangle\|\varphi\|^{2}\right] Y \\ &+[(X \varphi)\langle Y, Z\rangle-(Y \varphi)\langle X, Z\rangle] \nabla \varphi \end{aligned} $$
where $$ {H}_{\psi}(X) \doteq \nabla_{X}(\nabla \psi) $$
I don't understand how the equalities above outlined in red follow from $(1.2)$. Using $(1.2)$, I get:
$$ \begin{aligned} \overline{R}(X, Y) Y &=R_{m}(X, Y) Y-\left[\langle X, Y\rangle H_{\varphi} Y-\langle Y, Y\rangle H_{\varphi} X\right] \\ &+\left[\nabla^{2} \varphi(Y, Y)+(Y \varphi)(Y \varphi)-\langle Y, Y\rangle\|\varphi\|^{2}\right] X \\ &-\left[\nabla^{2} \varphi(X, Y)+(X \varphi)(Y \varphi)-\langle X, Y\rangle\|\varphi\|^{2}\right] Y \\ &+[(X \varphi)\langle Y, Y\rangle-(Y \varphi)\langle X, Y\rangle] \nabla \varphi \end{aligned} $$
so that (using the fact $X, Y$ are orthonormal) we have:
$$ \begin{aligned} \psi^{-2} \overline{K}_{\sigma} &=\langle \overline{{R}}(X, Y) Y, X \rangle\\ &={K}+ \langle H_{\varphi}X, X \rangle + \nabla^{2} \varphi(Y, Y)+(Y \varphi)^{2} -\|\nabla \varphi\|^{2} + \langle (X\varphi)\nabla \varphi, X \rangle\\ &=K+\nabla^2 \varphi(X, X) +\nabla^2 \varphi(Y, Y) + (Y\varphi)^2 + \langle (X\varphi)\nabla \varphi, X \rangle-\|\nabla \varphi\|^{2} \end{aligned} $$
so my doubt is really this one: how/why are the following equalities true:
$$\langle (X\varphi)\nabla \varphi, X \rangle = (X\varphi)^2$$ $$\nabla^2 \varphi(X, X) +\nabla^2 \varphi(Y, Y) + (Y\varphi)^2 + (X\varphi)^2 = \frac{2}{n} \cdot \frac{\Delta \psi}{\psi} $$
EDIT: I'm being silly. The first equality is by definition. The second one is still not solved though.
The first thing to do is to find a relation between $\nabla^2\psi$ and $\nabla^2\varphi$. Since $\psi = e^\varphi$, apply ${\rm d}$ to get ${\rm d}\psi = e^{\varphi}{\rm d}\varphi =\psi \,{\rm d}\varphi$. Transform both sides into vector fields using $g$ to get $\nabla\psi$ = $\psi \nabla \varphi$. Thus $$\nabla_X(\nabla\psi)=\nabla_X(\psi\nabla\varphi)=(X\psi)\nabla\varphi +\psi\nabla_X(\nabla\varphi),$$so take the $g$-product with $Y$ and get $$\nabla^2\psi(X,Y) = (X\psi)(Y\varphi)+\psi\nabla^2\varphi(X,Y) = \psi(X\varphi)(Y\varphi)+\psi\nabla^2\varphi(X,Y).$$
Hence $$\begin{align} \cdots &=K+\nabla^2\varphi(Y,Y)+\nabla^2\varphi(X,X)+(Y\varphi)^2+(X\varphi)^2-\|\nabla\varphi\|^2 \\ &= K +\frac{1}{\psi}\nabla^2\psi(X,X) +\frac{1}{\psi}\nabla^2\psi(Y,Y)- \|\nabla\varphi\|^2 \\ &\stackrel{(3)}{=} K +\frac{1}{\psi}\frac{\triangle\psi}{n} +\frac{1}{\psi}\frac{\triangle\psi}{n} -\|\nabla\varphi\|^2 \\ &= K + \frac{2}{n}\frac{\triangle\psi}{\psi}-\|\nabla\varphi\|^2.\end{align}$$