<s>[INST] How find $I= \int_{x=0}^{ \frac{1}{2} } \int_{y=x}^{1-x} ( \frac{x-y}{x+y})^{2}\, dy\,dx$ In $$I= \int_{x=0}^{ \frac{1}{2} }  \int_{y=x}^{1-x}  \left( \frac{x-y}{x+y}\right)^{2} \,dy\,dx$$
follow the change of variables on $x= \frac{1}{2} (r-s),y= \frac{1}{2} (r+s)$ and find$I$
My try
$J=\begin{vmatrix} x'_{r} & x'_{s}\\ y'_{r} & y'_{s} \end{vmatrix}=\begin{vmatrix} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{vmatrix}=\frac{1}{2}$
$x-y=\frac{1}{2}r-\frac{1}{2}s-\frac{1}{2}r-\frac{1}{2}s=-s$
$x+y=\frac{1}{2}r-\frac{1}{2}s+\frac{1}{2}r+\frac{1}{2}s=r$
and what next?
 [/INST] Off to a good start.  Draw the picture!  The hard part is determining the integration limits.  Really, you just need to understand what the lines of constant $r$ and $s$ mean.  Basically, they are lines at 45-degree angles to the axes.  So, really, imagine that if $r \in [0,1]$, then what are the limits of $s$?  Well, in fact, $s$ ranges from $0$ at the origin, to its maximum value at the top of the triangle on the $y$ axis, or $r$.  Thus, the integral is equal to
$$\frac12 \int_0^1 dr \, \int_0^r ds \, \left (\frac{s}{r} \right )^2 $$
This is easily doable.  I get $1/12$.
</s>