Show that a definite integral vanishes for all values of $a,b,c > 0$.

The result is surprising because one would think that the combination with the exponentials would be too complicated to make the integral vanish for all $a,b,c$. However, the exponentials don't enter into it because the integral along every line with constant sum $x+y$ is zero, and the exponential factor is constant on these lines. Consider

$$ x=u+v\;, \\ y=u-v\;, $$

which, up to a constant factor from the Jacobian, transforms the integral into

$$\int_0^\infty\mathrm du\mathrm e^{-2u}\int_{-u}^u\mathrm dv\frac{a(u+v)-b(u-v)}{\left(a^2(u+v)+b^2(u-v)+c(u^2-v^2)\right)^{3/2}}\;.$$

The inner integral is readily performed; Wolfram|Alpha gives

$$\int_0^\infty\mathrm du\mathrm e^{-2u}\frac2{(a-b)^2+2cu}\left[\frac{a(u+v)+b(u-v)}{\sqrt{a^2(u+v)+b^2(u-v)+c(u^2-v^2)}}\right]_{v=-u}^{v=u}\;,$$

and the result is $0$ since the antiderivative takes the same value at $u$ and $-u$.

Using the substitution $$ r\frac{1+t}{1-t}=\frac1r\frac{1+s}{1-s} $$ maps $[-1,1]\mapsto[-1,1]$ and has proven useful in several situations similar to this. Some formulas pertinent to this substitution are $$ \frac{\mathrm{d}t}{1-t^2}=\frac{\mathrm{d}s}{1-s^2} $$ $$ \frac{r(1+t)+\frac1r(1-t)}{1-t^2}=\frac{\frac1r(1+s)+r(1-s)}{1-s^2} $$ $$ \frac{r(1+t)-(1-t)}{\sqrt{1-t^2}}=\frac{(1+s)-r(1-s)}{\sqrt{1-s^2}} $$ Change variables $u=x+y$ and $v=x-y$, then $v=ut$: $$ \begin{align} &\int_0^\infty\int_0^\infty\frac{(ax-by)\,\mathrm{e}^{-x}\,\mathrm{e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}}\,\mathrm{d}x\,\mathrm{d}y\\[6pt] &=2\int_0^\infty\int_{-u}^u\frac{(a-b)u+(a+b)v}{\left(cu^2-cv^2+2(a^2+b^2)u+2(a^2-b^2)v\right)^{3/2}}\,\mathrm{e}^{-u}\,\mathrm{d}v\,\mathrm{d}u\\[6pt] &=2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)t}{\left(cu^2(1-t^2)+2u(a^2(1+t)+b^2(1-t))\right)^{3/2}}\,\mathrm{d}t\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{1} \end{align} $$ Using the substitution $\frac{a}{b}\frac{1+t}{1-t}=\frac{b}{a}\frac{1+s}{1-s}$ with $(1)$ yields $$ 2\int_0^\infty\int_{-1}^1\frac{(b-a)+(a+b)s}{\left(cu^2(1-s^2)+2u(b^2(1+s)+a^2(1-s))\right)^{3/2}}\,\mathrm{d}s\,\mathrm{e}^{-u}u^2\,\mathrm{d}u $$ Then substituting $w=-s$ yields $$ -2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)w}{\left(cu^2(1-s^2)+2u(a^2(1+w)+b^2(1-w))\right)^{3/2}}\,\mathrm{d}w\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{2} $$ Since $(1)$ and $(2)$ are equal, yet negatives, they are both $0$.

Take $a=c=1$ and $b$ very small.