Help with Change of Variable for the function $f(x,y)=e^{\frac{x}{2x+3y}}$ Let $D$ be the open triangle with the vertices $(0,0), (3,0), (0,2)$. For $f(x,y)=e^{ \frac{x}{2x+3y}}$ show that $f$ is integrable on $D$ and prove that $\iint_Df(x,y)dxdy=6\sqrt{e}-6$.
I was able to prove that $f$ is integrable on $D$, since $f$ is continuous everywhere but $(0,0)$ and around $(0,0)$, we have that $\frac{x}{2x+3y}<\frac{x}{2x}=\frac{1}{2}$, and therefore $f$ is bounded with a finite number of points where it's not continuous, and therefore, is integrable.
I also represented $D$ as $x\in (0,3)$ , $ 
y\in (0,-\frac{3x}{2}+3)$ since $y=-\frac{3x}{2}+2 $ is the hypertenuse of $D$. 
Once I got to calculating the integral itself, I tried multiple changes of variables, such as polar, $u=x, v=\frac{1}{2x+3y}$, $u=x, v=2x+3y$, $u=\frac{1}{2x+3y}, v=-\frac{3x}{2}+3$, and none of these gave an integral that could be calculated using analytical tools only (no numerical tools. I also checked this with mathematica and all of these integrals require numerical tools to calculate).
What change of variables can be used here? Thanks!

I think you had the upper limit for $y$ the wrong way around, unless you switched the vertices on the axes. If the vertices are on $(3,0)$ and $(0,2)$, the equation of the line joining them is $2x+3y=6$ so $y$ runs from $0$ to $-\tfrac{2}{3}x+2$.
If you let $u=x$ and $v=2x+3y$, then $u$ keeps the limits of $x$ and $v$ will go from $2u$ to $6$. The inverse relations are $x=u$ and $y=\tfrac{v}{3}-\tfrac{2u}{3}$, so the Jacobian is:
$$\begin{vmatrix}
1 & 0 \\
-\tfrac{2}{3} & \tfrac{1}{3}
\end{vmatrix} = \frac{1}{3}$$
The integral becomes:
$$\int_0^3 \int_0^{-\tfrac{2}{3}x+2} e^{\frac{x}{2x+3y}}\,\mbox{d}y \,\mbox{d}x = \frac{1}{3} \int_0^3 \int_{2u}^{6} e^{\frac{u}{v}}\,\mbox{d}v \,\mbox{d}u = (*)$$
Now $e^{\frac{u}{v}}$ doesn't have an elementary anti-derivative w.r.t. $v$, but you can change the order of integration. In the $uv$-plane, with $u$ from $0$ to $3$ and $v$ from $2u$ to $6$, the region is the triangle with vertices $(0,0)$, $(0,6)$ and $(3,6)$. Letting $v$ run fixed from $0$ to $6$ then gives limits for $u$ running from $0$ to $\tfrac{v}{2}$. The integral becomes easy to compute:
$$\begin{array}{rcl}
\displaystyle (*) = \frac{1}{3} \int_0^6 \int_{0}^{\tfrac{v}{2}} e^{\frac{u}{v}}\,\mbox{d}u \,\mbox{d}v
& = & \displaystyle \frac{1}{3} \int_0^6 \left[ ve^{\frac{u}{v}} \right]_{u=0}^{u=\tfrac{v}{2}} \,\mbox{d}v \\[8pt]
& = & \displaystyle \frac{1}{3} \int_0^6 \left( \sqrt{e}-1 \right)v \,\mbox{d}v \\[8pt]
& = & \displaystyle \frac{1}{3} \left( \sqrt{e}-1 \right) \left[ \frac{v^2}{2} \right]_{v=0}^{v=6} \\[8pt]
& = & \displaystyle 6\left( \sqrt{e}-1 \right) 
\end{array}$$
