= Solution
Invert the change of variables: $x=(v-u)/2$, $y=(v+u)/2$. The absolute <Jacobian determinant> is $1/2$, and the transformed triangle is $0\le v\le3$, $-v\le u\le v$. Hence
$$
I=\frac12\int_0^3\int_{-v}^{v}e^{u/v}\,du\,dv
=\frac12\int_0^3v(e-e^{-1})\,dv
=\boxed{\frac94(e-e^{-1})=\frac92\sinh1.}
$$
At the single vertex $v=0$, the printed integrand is undefined. This causes no integral divergence: throughout the remaining triangle $-1\le u/v\le1$, so the integrand is bounded between $e^{-1}$ and $e$. Giving the vertex any finite value, or taking a limiting integral with $v>0$, gives the same answer.
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