Proving that $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$ does not exist over $0 \leq x \leq 1 , 0 \leq y \leq 1$ To prove that double integral does not exist:     

$$\iint\frac{x-y}{(x+y)^2}\,dx\,dy$$ over region $0 \leq x \leq 1 , 0 \leq y \leq 1$.

I put $x - y = u$ and $x+y = v$ and I got integral as 
$$\iint \frac{u}{v^3}\,du\,dv$$ 
Limits of $u$ are from $2-v \leq u\leq v-2$ and $v$ are from $0$ to $1$. I am not sure about my new limits and how to prove it further.
EDIT : The correct expression is $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$
 A: There are four interpretations of this question, which explains seemingly conflicting answers.
(1) This does not exist as a Riemann integral.  For a function to be Riemann integrable it must be bounded, but we have 
$$\lim_{x \to 0+, y = 0} \frac{x-y}{(x+y)^3} = \lim_{x \to 0+} \frac{1}{x^2} = +\infty$$
An unbounded function $f:(0,1]^2 \to \mathbb{R}$ cannot satisfy the definition of Riemann integrability. For partitions $P$ of the domain into subrectangles, the limit of Riemann sums $S(P,f)$ as the partition norm $\|P\| \to 0$ will fail to exist.
(2) The integrand is also not Lebesgue integrable since it fails to be absolutely integrable.  Using polar coordinates, for any $\delta > 0$, we have
$$\begin{align}\int_{(0,1]^2} \left|\frac{x-y}{(x+y)^3} \right| &= \int_{(0,1]^2} \frac{|x-y|}{(x+y)^3} \\  &\geqslant \int_0^{\pi/2}\int_\delta^1 \frac{r |\cos \theta - \sin \theta|}{r^3(\cos \theta + \sin \theta)^3} r \, dr \, d\theta \\ &= \int_\delta^1 \frac{dr}{r}\int_0^{\pi/2} \frac{ |\cos \theta - \sin \theta|}{(\cos \theta + \sin \theta)^3} \, d \theta \\ &= -\log \delta \int_0^{\pi/2} \frac{ |\cos \theta - \sin \theta|}{(\cos \theta + \sin \theta)^3} \, d \theta  \\ &= -0.5 \log \delta \end{align},$$
and the RHS converges to $+\infty$ as $\delta \to 0+$.
(3) As an iterated integral, by antisymmetry it is clear that 
$$\int_0^1 \left(\int_0^1 \frac{x-y}{(x+y)^3} \, dx\right)\, dy = -1 \cdot \int_0^1 \left(\int_0^1 \frac{x-y}{(x+y)^3} \, dy\right)\, dx, $$
and depending on the order of integration the value is $\pm \frac{1}{2}$. Inequality of the iterated integrals also implies non-integrability both in the sense of Riemann and Lebesgue, since, otherwise, Fubini’s theorem would be violated.
(4) As an improper integral the principal value obtained by integrating over $S_\delta = [0,1]^2 \setminus \{(x,y): \sqrt{x^2+y^2} = \delta, x \geqslant 0, y \geqslant 0\}$ is
$$\lim_{\delta \to 0+} \int_{S_\delta} \frac{x-y}{(x+y)^3} = 0$$
A: Given Integral $$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^2} dx dy \\
                                  & = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^2} dx dy \\ & = \int^1_0 \int^1_0 \cfrac{1}{(x+y)} dx dy - 2 \int^1_0 \int^1_0 \cfrac{y}{(x+y)^2} dx dy \\ & = \int^1_0 \ln(x+y) \bigg|_0^1 dy + 2 \int^1_0 \cfrac{y}{x+y} \bigg |^1_0 dy  \\ & = \int^1_0\ln (y+1) dy - \int^1_0\ln(y) dy   - 2 \int^1_0 \cfrac{dy}{y+1} \\& = (y+1)(\ln(y+1)-1  )\bigg|_0^1 - y(\ln(y)-1)  \bigg|_0^1 - 2 \ln(y+1) \bigg |^1_0 \end{align}$$
Which is easy to evaluate, and I think the only slight problem would be in evaluating $y \ln(y) $ at $x = 0$, which is just $0$.
I don't see any problem with this.

After changing the question
Given integral:
$$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dx dy \\
                                  & = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^3} dx dy  \\& = \int^1_0 \int^1_0 \cfrac{dx dy}{(x+y)^2} - 2 \int^1_0 \int_0^1 \cfrac{y}{(x+y)^3} dx dy \\& = -\int^1_0  \cfrac{ dy}{(x+y)} \bigg |^1_0 + \cfrac{2}{2} \int^1_0  \cfrac{y}{(x+y)^2} \bigg |^1_0 dy  \\& = - \int^1_0 \bigg( \cfrac{1}{y+1} - \cfrac{1}{y} \bigg) dy + \int^1_0 \bigg( \cfrac{y}{(1+y)^2} - \cfrac{1}{y} \bigg ) dy \\ & = - \int^1_0 \cfrac{dy}{(1+y)^2}  \\& = {- \cfrac{1}{2}}\end{align}$$
But $$ \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dy dx = \cfrac{1}{2} $$
Then from converse of Fubini's theorem this integral doesn't exists.
A: This integral evaluates to $0$. Expanding on Ajay's answer, working out those bounds gives exactly $0$.
$$2(\mathrm{ln}(2)-1)+1+1-0-2\mathrm{ln}(2)-0 = 0$$
Solving the integral as attempted in the problem, one gets:
\begin{align*}
\int_0^2\int_{2-v}^{v-2}\frac{u}{v^3}\,du\,dv &= \int_0^2\frac{1}{v^3}\left(\frac{1}{2}u^2\bigg|_{2-v}^{v-2}\right)\\
&= \int_0^2\frac{1}{v^3}0\,dv = 0
\end{align*}
