= Solution
Fix $y\ne0$. The <function> $x\mapsto f(x,y)$ is continuous on $[-1,1]$, because its denominator is bounded away from zero, and it is an <odd function> of $x$. Its <integral> over this symmetric <interval> is therefore zero. At $y=0$ the entire section is zero, including the specified value at the origin. Thus the inner $x$ <integral> exists for every $y$ and is identically zero. The same reasoning with $x$ and $y$ exchanged gives
$$
\boxed{\int_{-1}^1\left(\int_{-1}^1f(x,y)\,dx\right)dy
=\int_{-1}^1\left(\int_{-1}^1f(x,y)\,dy\right)dx=0.}
$$
To test <Lebesgue integrability> in two dimensions, examine the <absolute value> near the origin. For $0<\varepsilon<1$, the annulus $\varepsilon<\sqrt{x^2+y^2}<1$ lies inside the square. In <polar coordinates>,
$$
|f(r\cos\theta,r\sin\theta)|=\frac{|\cos\theta\sin\theta|}{r^4},\qquad dx\,dy=r\,dr\,d\theta.
$$
The <change of variables formula> on this annulus, where the <integrand> has no singularity, gives
$$
\begin{aligned}
\int_{[-1,1]^2}|f(x,y)|\,dx\,dy
&\geq\int_\varepsilon^1r^{-3}\,dr\int_0^{2\pi}|\cos\theta\sin\theta|\,d\theta\\
&=\frac{\varepsilon^{-2}-1}{2}\cdot2
=\varepsilon^{-2}-1.
\end{aligned}
$$
The lower bound tends to infinity as $\varepsilon\downarrow0$. Hence
$$
\boxed{\int_{[-1,1]^2}|f|=\infty,\qquad f\notin L^1([-1,1]^2).}
$$
Moreover reflection in either coordinate axis exchanges the <positive part> and <negative part> while preserving <Lebesgue measure>. Both parts therefore have infinite <integral>, so a signed double <Lebesgue integral> would involve the undefined expression $\infty-\infty$.
There is no contradiction with <Fubini's theorem>: its <absolute integrability> hypothesis is missing. This is an example of <finite iterated integrals without joint integrability>, with every individual one-dimensional section integrable despite the two-dimensional singularity.
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