To show that it is convex, it would be easier if you show that the Hessian of the function is positive semi-definite. In fact, the Hessian is
$$\begin{bmatrix} {2 \over x_1^3 x_2} & {1 \over x_1^2 x_2^2} \\ {1 \over x_1^2x_2^2} & {2 \over x_1 x_3^3} \end{bmatrix}$$
The matrix and all of its principal submatrices are positive semi-definite, so it is convex, hence quasiconvex as well.
Or you can show PSD by showing $z^T H z \ge 0$, and we get
$$ {2 \over x_1^3 x_2}z_1^2 + {1 \over x_1^2 x_2^2}z_2z_1 + {1 \over x_1^2x_2^2}z_1z_2 + {2 \over x_1 x_3^3}z_2^2 \ge 0$$
The above is non-negative since you can complete the square.
I actually am not familiar with the method you're using, so I'm using a different method.
The function is actually concave on the (natural) domain $\Bbb{R} \times (0, \infty)$. Note that the function $(x_1, x_2) \mapsto \ln(x_2)$ is concave, because the function $\ln$ is concave (check its second derivative). The function $(x_1, x_2) \mapsto x_1$ is an affine function, and hence is concave (and convex). Summing two concave functions produces a concave function, and every concave function is quasiconcave.
The only question remaining is strictness. Neither of the above functions are strictly (quasi)concave, so we need a separate argument. Strict quasiconcavity means that, for all $(x_1, x_2), (y_1, y_2)$ in the domain, and any $\lambda \in (0, 1)$, we have
$$f(\lambda x_1 + (1 - \lambda)y_1, \lambda x_2 + (1 - \lambda)y_2) > \min \{f(x_1, x_2), f(y_1, y_2)\}.$$
Quasiconvexity means the above with $\ge $ substituted for $>$. So, let's suppose that we have equality. Without loss of generality, assume $f(x_1, x_2) \le f(y_1, y_2)$. Then,
$$\lambda x_1 + (1 - \lambda)y_1 + \ln(\lambda x_2 + (1 - \lambda)y_2) = x_1 + \ln(x_2).$$
By the strict concavity of $\ln$ (again, examine the second derivative), we have
\begin{align*}
&x_1 + \ln(x_2) > \lambda x_1 + (1 - \lambda)y_1 + \lambda\ln(x_2) + (1 - \lambda)\ln(y_2) \\
\iff \, &(1 - \lambda)x_1 + (1 - \lambda)\ln(x_2) > (1 - \lambda)y_1 + (1 - \lambda)\ln(y_2) \\
\iff \, &x_1 + \ln(x_2) > y_1 + \ln(y_2) \iff f(x_1, x_2) > f(y_1, y_2),
\end{align*}
which contradicts $f(x_1, x_2) \le f(y_1, y_2)$. Therefore, $f$ is quasiconcave.
Best Answer
Consider $A_\alpha = \{(x_1, x_2) \in \mathbb{R}_+^2: (x_1 + \gamma)x_2 \ge \alpha\}$.
If $\alpha \le 0$, then $A_\alpha = \mathbb{R}_+^2$ which is convex but not strictly convex.
If $\alpha > 0$, then $A_{\alpha} = \{(x_1, x_2)\in \mathbb{R}_+^2:x_2 \ge \frac{\alpha}{x_1+\gamma}\}$ which is convex.
Hence $u$ is quasi-concave.
It is easay to check that it is not concave by examining the Hessian.