$~D:=$domain where it is bounded and closed in $~ \mathbb R^2 ~$ and line symmetric with $~ y=x ~$ $$ I:=\underbrace{\iint_{D} {x^2-y^2 \over 1+x^4+y^4 } \mathrm{d}x \mathrm{d}y}_{\text{I want to evaluate this double integral} } $$
I guess that $~D~$ is a domain of, for all $~(a,a)~$ on $~y=x~$,euclid distances between each boundary of $D_1,D_2$ and $y|_{x=a}=x|_{x=a}$ are same.
I've written the example of the domain $~D~$ as follwoing.
ADDed
As the advice from @Hans Lundmark says, seems at least, $I$ takes a constant value such as $0$ .
$$ I:=\iint_{D} {x^2-y^2 \over 1+x^4+y^4 } \mathrm{d}x \mathrm{d}y $$ Swap the names $x$ and $y$ then $$ I=\iint_{D} {y^2-x^2 \over 1+x^4+y^4 } \mathrm{d}y \mathrm{d}x $$ Then notice that $$ -I=\iint_{D} {x^2-y^2 \over 1+x^4+y^4 } \mathrm{d}y \mathrm{d}x $$ Since the region $D$ is symmetric if $x$ and $y$ are actually swapped, we can swap the region of integration, so $$ -I=\iint_{D} {x^2-y^2 \over 1+x^4+y^4 } \mathrm{d}x \mathrm{d}y=I $$ so we have $-I=I$, the only solution of which is that $I=0$.
Intuitively, if the region $D$ is symmetric with respect to reflection in the line $y=x$, then for each $y^2-x^2$ in the integrand, we have an equivalent numerator $x^2-y^2$ with the same denominator on the other side of the line $y=x$ which cancels it out.
Another way to see this is to change variables to $u=x+y$ and $v=x-y$. You then get an integral like $$ \int_{R}\int_{-v_0}^{v_0} {uv\over 1+(u^2+v^2)^2/4-(u^2-v^2)^2/16}dv \;du $$ which is an odd function of $v$, so you can see that it integrates to zero identically. Here we have the limits of integration from $-v_0$ to $v_0$ because our region of integration is symmetric around the line $y=x$. This is somewhat like rotating our "axes" by $45^\circ$ anticlockwise.
(Here $(u^2+v^2)^2=2(x^2+y^2)$ and $u^2-v^2=4xy$ so $4(u^2+v^2)^2-(u^2-v^2)^2=16(x^4+y^4)$.)