Suppose $\gamma \in R^{1}$ and $\beta \in R^{k}$.
Let $f(\gamma,\beta) = (y_{2} - \gamma y_{1}) - (y_{3} - \gamma y_{2}) \exp(x^{\prime}\beta)$.
Then is $f$ a concave function of $(\gamma,\beta^{\prime})$? I just checked its Hessian. It seems the Hessian is negative semi-definite.
Computing the Hessian of $$ (y_2-\gamma y_1)-(y_3-\gamma y_2)e^{x\cdot\beta} $$ we get $$ \begin{align} H &=\frac{\partial\nabla f}{\partial(\gamma,\beta)}\\[12pt] &=\frac{\partial\left(-y_1+y_2e^{x\cdot\beta},(\gamma y_2-y_3)xe^{x\cdot\beta}\right)}{\partial(\gamma,\beta)}\\[6pt] &=\begin{bmatrix}0&y_2x\\[3pt]y_2x'&(\gamma y_2-y_3)x'x\end{bmatrix}e^{x\cdot\beta} \end{align} $$ Therefore, $$ \begin{align} e^{-x\cdot\beta}\begin{bmatrix}u&v\end{bmatrix}H\begin{bmatrix}u\\v'\end{bmatrix} &=\begin{bmatrix}y_2v\cdot x&\left(y_2u+(\gamma y_2-y_3)v\cdot x\right)x\end{bmatrix}\cdot\begin{bmatrix}u&v\end{bmatrix}\\ &=y_2uv\cdot x+\left(y_2u+(\gamma y_2-y_3)v\cdot x\right)v\cdot x\\[6pt] &=2y_2uv\cdot x+(\gamma y_2-y_3)(v\cdot x)^2 \end{align} $$ If $y_2=0$ and $y_3\le0$, then the Hessian would be positive.