I have been told to investigate the existence and equality of the integrals; $\int_{[0,1]^2} f\;d\lambda^2$, $\int_0^1\int_0^1 f\;d\lambda(x)d\lambda(y)$ and $\int_0^1\int_0^1 f\;d\lambda(y)d\lambda(x)$ for $f(x,y)=(x^2-y^2)(x^2+y^2)^{-2}$
Here is my answer, I spent a long time thinking about it with a friend,
f is cts on $(0,1)^2$ and since $\lambda_2(\partial((0,1)^2))=0$ it follows that f is cts almost everywhere and hence f is $\lambda_2$-measurable so $\int_{[0,1]^2} f\;d\lambda^2$ exists.
Now evaluating the other two as if they were Riemann integrals gives one as $\pi/4$ and the other as $-\pi/4$ so by the contrapositive of Fubinis theorem as given on wiki we have $\int_{[0,1]^2} |f|\;d\lambda^2=\int_{[0,1]^2} f^+\;d\lambda^2+\int_{[0,1]^2} f^-\;d\lambda^2=\infty$ which implies $\int_{[0,1]^2} f\;d\lambda^2=\pm\infty$ so all exist but none are equal.
I don't know whether you can evaluate the latter integrals in the question as Riemann integrals seeing as f is not bounded on the space, I also don't know whether my first argument actually shows f is $\lambda_2$-measurable