$\iint (y^2-x^2) e^{xy} dxdy$ $\iint_G(y^2-x^2)e^{xy}dxdy$ on G = $\{ (x,y) | x \ge 0, 1 \le xy \le 4, 0 \le y-x \le 3 \}$
I have tried a substitution of 
$u = xy$
$v = y-x$
which got me to $\bar G = \{ (u,v) | 1 \le u \le 4, 0 \le v \le 3 \}$
and
$\iint_\bar G \left(\frac{v}{\sqrt{v^2+4u}}\right)^{\frac12} e^u dudv$
(note that $x = -\frac12v+\frac12\sqrt{v^2+4u}$
and that $y = \frac12v+\frac12\sqrt{v^2+4u}$ ) 
But from here on i'm stuck.

A slightly different change of variables which simplifies a lot the boundary conditions :
$\begin{cases}
    s=y-x\\
    p=xy\\
  \end{cases}$
The Jacobian is :  $-\begin{vmatrix}
  -1 & 1 \\
  y & x \\ 
 \end{vmatrix}^{-1}=\frac{1}{y+x}$
$(y^2-x^2)\frac{1}{y+x}=y-x=s$
$\iint_{1 \le xy \le 4, 0 \le y-x \le 3}(y^2-x^2)e^{xy}dxdy =
\int_1^4\int_0^3 s e^p ds\;dp=
\int_1^4 e^p dp\int_0^3 s ds=
(e^4-e)\frac{3^2}{2}$ 
