Proof of violation of hypothesis Fubini-Tonelli's theorem

Question

Consider $f=\frac{8xy(x^2-y^2)}{(x^2+y^2)^3}$ for $(x,y) \neq (0,0)$, and Lebesgue measure on $[0,1]$, so the measure space is $[0,1]\times[0,1]$. I want to show that the hypotheses of Fubini-Tonelli's theorem are violated. First, I show: $$ f^{+} = \max(f, 0) \ \text{ for } 0<y<x<1\\ f^{-} = \max(-f, 0) \ \text{ for } 0<x<y<1 $$ For the theorem to be valid, either (or both) must be true: $$ \int_{[0,1]\times [0,1]}f^{+}<\infty\\ \int_{[0,1]\times [0,1]}f^{-}<\infty $$ First, for $f^{+}$, I switch to polar coordinates (since $\sin \theta <\cos \theta$ for $0<\theta<\frac{\pi}{4}$): $$ \int_{0}^{1}\int_{0}^{x}f^{+}\lambda(dy)\mu(dx) = \int_{0}^{1}\int_{0}^{\frac{\pi}{4}}\frac{8r^2 \cos \theta \sin \theta(\cos^2 \theta - \sin^2 \theta)r dr d\theta}{r^6} = \\\int_{0}^{1}\int_{0}^{\frac{\pi}{4}} \frac{4 \sin 2 \theta \cos 2 \theta dr d\theta}{r}$$ The inner integral evaluates to $1$, and the second one diverges. Therefore, $$ \int_{[0,1]\times[0,1]}f^{+} = \infty $$ Performing similar manipulations with $f^{-}$, $$ \int_{0}^{1}\int_{x}^{1}f^{-}\lambda(dy)\mu(dx) = \int_{0}^{1}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{4 \sin 2 \theta \cos 2 \theta dr d\theta}{r} $$ The inner integral evaluates to $-1$, and the second one diverges. Therefore, $$ \int_{[0,1]\times[0,1]}f^{-} = -\infty $$ So in total I get the expression $$ \int f = \int f^{+} + \int f^{-} = \infty - \infty $$ which violated the hypothesis of Fubini-Tonelli.