All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get
$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$
We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$.
Comment: If you find this not obvious, let's look at the first term, that is, at $\frac{ab+ca}{a^2+ab+ca}$ The "top" is $a(b+c)$. The "bottom" is $a(a+b+c)$. So the fraction is $\frac{a(b+c)}{a(a+b+c)}$ Divide top and bottom by $a$, or equivalently, "cancel" the $a$'s. We are using the "algebra" version of the familiar fact that $\frac{2\cdot 3}{2 \cdot 5}=\frac{3}{5}$
Additional comment The hint is strange. True, we can rewrite the expression as $\left(1-\frac{a^2}{a^2+ab+ca}\right) +\left(1-\frac{b^2}{b^2+ab+bc}\right)+\left(1-\frac{c^2}{c^2+bc+ca}\right)$ which is equivalent to what was given. And then we could divide top and bottom by $a$ in the first fraction, by $b$ in the second, by $c$ in the third, and end up with $3-\frac{a+b+c}{a+b+c}$ But it sure seems like a lot of work when we can cancel immediately! You are sure that you quoted the problem correctly?