We have been asked to investigate whether $\int_{[0,1]^2} f \, d\lambda^2$ exists and if so whether it is equal to $\int_0^1 (\int_0^1f(x,y) \, d\lambda(x)) \, d\lambda(y)$ and $\int_0^1 (\int_0^1 f(x,y) \, d\lambda(y)) \, d\lambda(x)$ for $f(x,y) = (x^2 - y^2)(x^2 + y^2)^{-2}$.
So far we have worked out that since $f$ is continuous on the unit square it is $\lambda_2$ lebesgue measurable. So for $f$ to satisfy Fubini's theorem we just need to show $\int_{[0,1]^2}|f| \, d\lambda^2$ is intergrable. We were wondering if it is enough to show that the Riemann integral $\int_0^1 (\int_0^1 f(x,y) \, dx) \, dy$ is integrable, and if this means the Lebesgue integral is equal to the Riemann integral?
If you compute the integrals $\int_0^1 \int_0^1 f(x,y)\, dx\, dy$ and $\int_0^1 \int_0^1 f(x,y)\, dy\, dx$ directly, you would find that they are not equal (the latter integral is $\pi/4$ and the former is $-\pi/4$). Furthermore, the integral $\int_{[0,1]^2} |f|\, d\lambda_2$ does not exist. For if it did, then since $|f|$ is nonnegative, Fubini's theorem would imply $\int_0^1 \int_0^1 |f(x,y)|\, dy\, dx < +\infty$. However,
\begin{align*}\int_0^1 \int_0^1 |f(x,y)|\, dy\, dx &= \int_0^1 \int_0^x \frac{x^2 - y^2}{(x^2 + y^2)^2}\, dy\, dx + \int_0^1 \int_x^1 \frac{y^2 - x^2}{(x^2 + y^2)^2}\, dy\, dx\\ &= \int_0^1 \left(\frac{y}{x^2 + y^2}\right)\bigg|_{y = 0}^{y = x} + \int_0^1 \left(-\frac{y}{x^2 + y^2}\right)\bigg|_{y = x}^{y = 1}\, dx\\ &= \int_0^1 \frac{1}{2x}\, dx + \int_0^1 \left(\frac1{2x} - \frac1{1 + x^2}\right)\, dx\end{align*}
and the last two integrals diverge to $+\infty$.