The $R$ is the region of integration, described in the following image.
To solve this integral we make a change of coordinates with $u=\sqrt{xy}$ and $v=\sqrt{\frac{y}{x}}$. Furthermore, in the region of integration we have $x>0$ and $y>0$. Then $u^2=xy$ y $v^2=\frac{y}{x}$, so we have: \begin{align*} v^2 &=\frac{y}{x}\\ x^2v^2 &= xy\\ x^2v^2 &= u^2. \end{align*} \begin{align*} x^2 &= \frac{u^2}{v^2}\\ x &= \frac{u}{v}. \end{align*} Observation: $x=-\frac{u}{v}$ is not taken, because $x>0$ in the $R$ region.
With the above we can solve for $y$. \begin{align*} v^2& = \frac{y}{x}\\ v^2& = \frac{y}{\frac{u}{v}}\\ uv & = y. \end{align*} On the other hand, in the $R$ region we have the following equations $xy = 1$, $y = x$ and $y = 2$. Now with all of the above, let's pass these equations to the coordinates $u$ and $v$. \begin{align*} xy = 1 &\Rightarrow u = \sqrt{1} \Rightarrow u = 1\\ y = x &\Rightarrow uv = \frac{u}{v} \Rightarrow v^2 = 1 \Rightarrow v = 1\\ y = 2 &\Rightarrow 2 = uv. \end{align*} Observation: By definition of $u$ and $v$, we have to be positive about the values taken by the $R$ region, therefore $v = -1$ is not taken.
Using the above equations, we have that the region $R$ is transformed into the region $S$, shown below.
Now using $x = \frac{u}{v}$ and $y = uv$ we calculate the Jacobian to use the change of variable in the original integral. \begin{align*} \frac{\partial x}{\partial u}&=\frac{1}{v} &\frac{\partial y}{\partial u} =v\\ \frac{\partial x}{\partial v}&=-\frac{u}{v^2} &\frac{\partial y}{\partial v} =u \end{align*} Thus, the Jacobian is: \begin{align*} \begin{vmatrix} \frac{1}{v} &v\\ -\frac{u}{v^2} &u \end{vmatrix} =\frac{u}{v}+\frac{u}{v}=\frac{2u}{v}. \end{align*} So using all the information developed we have: \begin{align*} \iint\limits_R\sqrt{\frac{y}{x}}e^{\sqrt{xy}}dA &= \iint\limits_Sve^u\left(\frac{2u}{v}\right)dA\\ &= 2\int_{1}^{2}\int_{1}^{2/u}ue^u\,dv\,du\\ &= 2\int_{1}^{2}ue^uv\biggr\rvert_{1}^{2/u}\,du\\ &= 2\int_{1}^{2}ue^u\left(\frac{2}{u}-1\right)du\\ &= 2\int_{1}^{2}e^u(2-u)du=2(e^u(3-u))\biggr\rvert_{1}^{2}=2e(e-2). \end{align*}
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
Your solution is correct. Still we can do without changing variables: \begin{align*} [\ldots]&=\int_1^2\int_{1/y}^y\sqrt\frac{y}{x}e^{\sqrt{xy}}\,dx\,dy \\&=\int_1^2 2e^{\sqrt{xy}}\Bigg|_{x=1/y}^{x=y}\,dy \\&=2\int_1^2(e^y-e)\,dy \\&=2(e^y-ey)\Bigg|_{y=1}^{y=2}=2e(e-2). \end{align*}