Let $f(x,y)=\frac{x-y}{(x+y)^3}$ and $g(x,y)=\text{sgn}(x-y)e^{-|x-y|}$ ( http://en.wikipedia.org/wiki/Sign_function )
We have $$\int_0^1\int_0^1f(x,y)dy dx = 1/2 = -\int_0^1\int_0^1f(x,y)dy dx$$ and $$\int_0^\infty\int_0^\infty g(x,y) dy dx = -\int_0^\infty\int_0^\infty g(x,y) dx dy=-1$$
Why doesn't Fubini theorem apply? For the first one, I think it's because of the pole at $(x,y)=(0,0)$ for $f(x,y)$. So that $f$ fails to be in $L^1 (\mathbb{R}^2)$. I'm not too sure about the second one.
What do you mean by "fail"? Are you asking why we can't apply Fubini's theorem, or what's the reason of the failure of the conclusion?
A version of the Fubini-Tonelli theorem states that if $f:\mathbb{R^2} \to \mathbb{R}$ is a measurable function, then $\int_{\mathbb{R}^2} f(x,y) d(x,y)$ may be computed as an interated integral (i.e., $\int_{\mathbb{R}^2} f(x,y) d(x,y) = \int_{\infty}^{\infty} (\int_{-\infty}^{\infty} f(x,y) dx )dy$) if one of the two conditions hold:
1) $f(x,y) \ge 0$ everywhere;
2) $\int_{\mathbb{R}^2} |f(x,y)| d(x,y) < \infty$
Combining 1) and 2), we see that we need that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |f(x,y)| dx dy < \infty$
So to show that Fubini's theorem cannot be applied, you need to show that...