Let $\Omega$ be an open subset of $\mathbb R^n$ . Let $\gamma\in C^1(\Omega)$ be bounded away from zero. Find $q,\beta\in C^1(\Omega)$ such that \begin{equation*} \nabla\cdot[\gamma\nabla u]=0\Leftrightarrow(-\Delta +q)v=0 \text{ for some }v=\beta u \end{equation*}
My attempt: $$ \begin{aligned} 0&=~ (-\Delta+q)\beta u\\ &=~-(u\Delta \beta+\beta\Delta u+2\nabla u\cdot\nabla \beta)+q\beta u\\ &=~-2\nabla u\cdot\nabla \beta+u(-\Delta \beta+q\beta)-\beta\Delta u \end{aligned}$$ Also we have $$ \begin{aligned} 0&=~ \nabla\cdot(\gamma\nabla u)\\ &=~\nabla u\cdot \nabla \gamma+\gamma \Delta u \end{aligned} $$
I do not know how should I proceed now.
There's something a little bit funny about the regularity assumptions on $\beta$ and $q$ if you're working with classical solutions since you're asked to evaluate $\Delta(\beta u)$ but only assume $\beta \in C^1(\Omega)$. I'm going to change the numerology to show the idea of the change of unknown while avoiding this issue. Indeed, I'll assume $\gamma \in C^2(\Omega)$.
Set $\beta = \gamma^{1/2}$, which belongs to $C^2(\Omega)$ since $\gamma \in C^2(\Omega)$ and is bounded away from $0$. Note that $\beta$ doesn't vanish, so we can define $q = \Delta \beta / \beta \in C^0(\Omega)$.
Then $$ \nabla \beta = \frac{1}{2} \gamma^{-1/2} \nabla \gamma $$ and $$ -\Delta \beta + \beta q =0. $$ Thus if the functions $u$ and $v$ are related by $v = \beta u$, then $$ (-\Delta + q) v = -\beta \Delta u - 2 \nabla u\cdot \nabla \beta + u(q\beta -\Delta \beta) = -\gamma^{1/2} \Delta u -\gamma^{-1/2} \nabla \gamma \cdot \nabla u \\ = -\gamma^{-1/2} \left(\gamma \Delta u + \nabla \gamma \cdot \nabla u \right) = -\gamma^{-1/2} \nabla \cdot (\gamma \nabla u). $$ We immediately deduce from this that $v$ satisfies $-\Delta v + qv =0$ if and only if $u$ satisfies $\nabla\cdot(\gamma \nabla u) =0$.