Answer
Please see the figure attached.
The iterated integral in the opposite order is
$\mathop \smallint \limits_1^{\rm{e}} \mathop \smallint \limits_0^{\ln y} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
Work Step by Step
We have $\mathop \smallint \limits_0^1 \mathop \smallint \limits_{{{\rm{e}}^x}}^{\rm{e}} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$.
From the limits of the integrals we obtain the domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,{{\rm{e}}^x} \le y \le {\rm{e}}} \right\}$
Notice that this is a vertically simple region.
Using this description we sketch the the domain of integration. Please see the figure attached.
We change the order of integration such that the domain becomes a horizontally simple region. In this case, the lower and upper boundaries are $y=1$ and $y = {\rm{e}}$, respectively. Whereas, the left boundary is $x=0$. To find the right boundary, we use $y = {{\rm{e}}^x}$ and get $x = \ln y$. Thus, the new domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|1 \le y \le {\rm{e}},0 \le x \le \ln y} \right\}$
So, the iterated integral in the opposite order is
$\mathop \smallint \limits_1^{\rm{e}} \mathop \smallint \limits_0^{\ln y} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
