Simple ain't it?

We have $ab+bc+ca=0 \\ \Rightarrow {bc=-(ab+ca)}\Rightarrow {-bc=ab+ac..........1} \\ \Rightarrow {ca=-(ab+bc)}\Rightarrow {-ca=ab+bc..........2} \\ \Rightarrow {ab=-(bc+ca)}\Rightarrow {-ab=bc+ca}..........3$

Now let's come to the problem, We have $\frac {1}{a^2-bc}+\frac {1}{b^2-ca}+\frac {1}{c^2-ab}\\ =\frac{1}{a^2+ab+ac}+\frac {1}{b^2+ba+bc}+\frac {1}{c^2+cb+ca} \\ =\frac {1}{a (a+b+c)}+\frac {1}{b (a+b+c)}+\frac {1 }{c (a+b+c)}\\=\frac {1}{a+b+c}\left (\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)\\=\frac {1}{a+b+c}\left (\frac {ab+bc+ca}{abc}\right)\\=\frac {1}{a+b+c}\times 0\\ \boxed{0}$

Bull's eye