Creating my own change of variables to evaluate an integral The question asks me to evaluate the integral $$\iint_{R} e^{\frac{x+y}{x-y}} dA$$ where $R$ is trapezoid region with the vertices $(1,0), (2,0), (0,-2), (0,-1)$. I'm supposed to suggest a possible transformation and integrate and sketch the two regions.
My work :
Let the transformation be
$u=x-y$, $v=x+y$
Then with some algebra, I get $x=\frac{u+v}{2}$, and $y=-\frac{1}{2} (u-v)$
$J(u,v)=\begin{vmatrix}
\frac{1}{2} & \frac{1}{2}\\ 
-\frac{1}{2} & \frac{1}{2}
\end{vmatrix}=\frac{1}{2}$
When I sketch the region I have something like this on the xy plane

On the uv plane the transformation looks like:

So the integral becomes
$$\int_{1}^{2}\int_{-u}^{u} e^{\frac{v}{u}}*\frac{1}{2} dv du$$
$$\frac{1}{2}\int_{1}^{2}u\Big(e-\frac{1}{e}\Big)du$$
$$=\frac{1}{2}\Big(e-\frac{1}{e}\Big)*\frac{3}{2}=\frac{3}{4}\Big(e-\frac{1}{e}\Big)$$
Does this look correct?

Your approach is nice and the calculation is well done. We can check it, by calculating it slightly different and look if the results coincide. We apply the identity
\begin{align*}
\frac{x+y}{x-y}=1+\frac{2y}{x-y}
\end{align*}
and consider
\begin{align*}
\iint_{R} e^{\frac{x+y}{x-y}} dA=e\iint_{R} e^{\frac{2y}{x-y}} dA\tag{1}
\end{align*}
We use the variable transformation
\begin{align*}
u&=y\qquad\qquad\qquad x=u+v\\
v&=x-y\qquad\qquad\ y=u
\end{align*}
The trapezoid regions have vertices
\begin{align*}
\mathrm{Tr}_{(x,y)}&=\{(0,-2),(0,-1),(1,0),(2,0)\}\\
\mathrm{Tr}_{(u,v)}&=\{(-2,2),(-1,1),(0,1),(0,2)\}\\
\end{align*}
and the transformed region is given by the graphic below.

The Jacobian determinant is
\begin{align*}
J(u,v)=
\begin{vmatrix}
x_u&x_v\\
y_y&y_v\\
\end{vmatrix}
=
\begin{vmatrix}
1&1\\
1&0
\end{vmatrix}=-1
\end{align*}

We obtain with the right-hand side of (1)
\begin{align*}
\color{blue}{e\iint_{R} e^{\frac{2y}{x-y}} dA}
&=e\int_1^2\int_{-v}^0e^{\frac{2u}{v}}|-1|\,du\,dv\\
&=e\int_{1}^2\left.\frac{v}{2}e^{\frac{2u}{v}}\right|_{-v}^0\,dv\\
&=\frac{e}{2}\int_{1}^2v\left(1-e^{-2}\right)\,dv\\
&=\frac{1}{2}\left(e-\frac{1}{e}\right)\left.\frac{1}{2}v^2\right|_{1}^2\\
&\,\,\color{blue}{=\frac{3}{4}\left(e-\frac{1}{e}\right)}
\end{align*}
in accordance with OPs result.

