When my mind isn't creative anymore...

Notice that $\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=3$ implies that, for $x=ab, y=bc, z=ca$ :

$x^3+y^3+z^3-3xyz=0\leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$ .

This implies that either $ab+bc+ca=0 \hspace{0.3cm}(1)$ or $a=b=c \hspace{0.3cm}(2)$ .

• Case (1):

We easily find, by applying (1), $\begin{aligned} &\frac{a+b}{c}=-\frac{ab}{c^2}\\ &\frac{c+b}{a}=-\frac{bc}{a^2}\\ &\frac{a+c}{b}=-\frac{ac}{b^2} \end{aligned}$

and we finally find that $\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=-(\frac{ab}{c^2} +\frac{bc}{a^2} +\frac{ac}{b^2})=-3$

by the original equation.

• Case(2): For $a=b=c$ it can be seen that

$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=6$

Hence the result follows. $\ _\square$