I have limited my graphs to two-dimensional ones, which will serve the purpose of illustrating the type of region is involved in this integration, since this is something you've asked about. (A three-dimensional plot turned out to be somewhat difficult to read.)
For triple integrals, a region is Type I if surfaces between which the integration is to be performed have a common "projection" onto the $ xy-$ plane $ \ ( z = 0 ) \ $. We would describe these surfaces using two functions $ \ z = f_1(x,y) \ $ and $ \ z = f_2(x,y) \ $ and the integration would be set up as $ \ \iint \int_{z_1}^{z_2} \phi(x,y,z) \ \ dz \ (dy \ dx) \ $ , the last two differentials being in either order.
For the integration in question here, we can look at the domains delimited on the planes $ \ z = 1 \ $ and $ \ z = 2 \ $ and find that their projections onto the $ \ xy-$ plane would differ. So this integration is not being conducted in a Type I region.
Correspondingly, a region is Type II if there is a common projections onto the $ yz-$ plane $ \ ( x = 0 ) \ $ , with the surfaces being two functions $ \ x = f_1(y,z) \ $ and $ \ x = f_2(y,z) \ $ and the integration being $ \ \iint \int_{x_1}^{x_2} \phi(x,y,z) \ \ dx \ (dy \ dz) \ $ . For a Type III region, the common projections are onto the $ xz-$ plane $ \ ( y = 0 ) \ $ , the surfaces are $ \ y = f_1(x,z) \ $ and $ \ y = f_2(x,z) \ $ , and the integration, $ \ \iint \int_{y_1}^{y_2} \phi(x,y,z) \ \ dy \ (dx \ dz) \ $ . [In practice, we don't generally worry about how these regions are labeled, but simply check to see how the integral may be arranged. As you might imagine, integration could occur over a three-dimensional region which falls into none of these categories.]
Returning to our integral again, we can look at the domains where the volume meets the $ \ y = 0 \ $ and $ \ y = 2 \ $ planes to see that we do not have a Type III region. However, the oblique plane $ \ x = y + z \ $ causes the volume to extend as far in the $ \ x-$ direction as $ \ x = 4 \ $ .
If we "sight down" the $ \ x-$ axis back toward the $ \ yz- \ $ plane, this oblique plane "approaches" us running left to right (toward increasing values of $ \ y \ $) and we find that there is a common projection onto the $ \ x = 0 \ $ plane, making the volume a Type II region. The bounding surfaces are thus the planes $ \ x = 0 \ $ and $ \ x = y + z \ $ . We would then set up and compute the triple integral
$$ \int_1^2 \int_0^z \int_0^{y+z} \ \frac{1}{(x + y + z)^3} \ \ dx \ dy \ dz \ \ = \ \ \int_1^2 \int_0^z \ \left[ \ -\frac{1}{2 \ (x + y + z)^2} \ \right] \vert_{x=0}^{x=y+z} \ \ dy \ dz $$
$$ = \ \ \frac{1}{2} \ \int_1^2 \int_0^z \ \left[ \ \frac{1}{ (0 + y + z)^2} \ - \ \frac{1}{ ([y+z] + y + z)^2} \right] \ \ dy \ dz $$
$$ = \ \ \frac{1}{2} \ \int_1^2 \int_0^z \ \frac{3}{ 4 \ (y + z)^2} \ \ dy \ dz \ \ = \ \ \frac{3}{8} \ \int_1^2 \ \left[ \ -\frac{1}{y + z} \ \right] \vert_{y=0}^{y=z} \ \ dz $$
$$ = \ \ \frac{3}{8} \ \int_1^2 \ \left[ \ \frac{1}{z} \ - \ \frac{1}{2z} \ \right] \ \ dz \ \ = \ \ \frac{3}{8} \ \int_1^2 \ \frac{1}{2z} \ \ dz \ \ = \ \ \frac{3}{16} \ \ln \ z \ \vert_1^2 \ = \ \frac{3}{16} \ln \ 2 \ \ . $$
Best Answer
To evaluate this integral most efficiently, it's better to switch to new variables that better reflect the region of integration (and won't ruin the integrand). Here's an outline of the solution.
Let $u=\frac{y}{x}$. Then two of the bounding equations, $y=\frac{x}{4}$ and $y=2x$ tell us that $u$ ranges in the interval $u\in\left[\frac{1}{4},2\right]$.
Let $v=xy$. Then the other two of the bounding equations, $y=\frac{1}{x}$ and $y=\frac{4}{x}$ tell us that $v$ ranges in the interval $v\in[1,4]$.
Now the given double integral can be rewritten as $$\iint_R e^{xy/2}\,dA=\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du,$$ where $J$ is the Jacobian of this change of variables.
Solving $u=\frac{y}{x}$ and $v=xy$ for $x$ and $y$, we find that $x=u^{-1/2}v^{1/2}$ and $y=u^{1/2}v^{1/2}$. Therefore the Jacobian is $$J=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} -\frac{1}{2}u^{-3/2}v^{1/2} & \frac{1}{2}u^{-1/2}v^{-1/2} \\ \frac{1}{2}u^{-1/2}v^{1/2} & \frac{1}{2}u^{1/2}v^{-1/2} \end{vmatrix}=-\frac{1}{2}u^{-1}.$$
Then the integral can be computed further as $$\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du=\int_{1/4}^2\int_1^4 e^{-v/2}\cdot\frac{1}{2}u^{-1}\,dv\,du=\frac{1}{2}\cdot\int_{1/4}^2 u^{-1}\,du\cdot\int_1^4 e^{-v/2}\,dv \\ =\frac{1}{2}\cdot\left[\ln|u|\right]_{1/4}^2\cdot\left[-2e^{-v/2}\right]_1^4=-\left(\ln2-\ln\frac{1}{4}\right)\cdot\left(e^{-2}-e^{-1/2}\right)=\left(e^{-1/2}-e^{-2}\right)\ln8.$$