Use the transformation $u = x + 2y$, $v = y-x$ to evaluate $\displaystyle \int_{0}^{\frac{2}{3}} \int_{y}^{2-2y}\left(x+2y\right)e^{y-x} \, dx \, dy$.
I started with calculating the jacobian:
$J(u,v) = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} = \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = 1(1) - 2(-1) = 1 + 2 = 3$
Next, I solved for an expression of $x$ and $y$ in tems of $u$ and $y$:
$$x = \frac{1}{3}(u - 2v)$$ $$y = \frac{1}{3}(u + v)$$
Now, I am stuck on how to find the new bounds of region $S$ for:
$$\iint_{S} \left(u\cdot e^{v}\right)\, dA$$
The limits of integration of the original integral, in terms of $(x,y)$ are given by $$ 0\leq y \leq \frac{2}{3}\,,\qquad y\leq x \leq 2-2y\,.$$
Rewriting with the proposed variables, we first have that \begin{align} 0\leq y \leq \frac{2}{3}&\Rightarrow 0\leq u+v \leq 2\,.\quad(1)\\ \end{align}
Then, \begin{align} y\leq x \leq 2-2y&\Rightarrow u+v\leq u-2v\leq 6-2(u+v) &\Rightarrow\begin{cases}v\leq 0\,.\quad(2)\\u\leq2\,.\quad(3)\end{cases} \\ \end{align}
From $(1)$, $(2)$ and $(3)$, we can rewrite $S$ in different forms. One option is given by
$$0\leq u\leq 2\,,\qquad-u\leq v\leq0\,,$$ that requires to integrate first by $v$ and then by $u$. Conversely, one can re-describe $S$ such that first we integrate by $u$ and then by $v$, for example $$-2\leq v\leq 0\,,\qquad-v\leq u\leq2\,.$$
Which one to choose? It does not really matter, but I'd choose the one that makes integration easier.