Now, just square it

$\begin{aligned} \frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} & = \left( \color{#3D99F6}{\frac 1{a} + \frac 1{b} + \frac 1{c}} \right)^2 - 2 \left( \frac 1{ab} + \frac 1{bc} + \frac 1{ca} \right) & \small \color{#3D99F6}{\text{Note that }\frac 1{a} + \frac 1{b} + \frac 1{c} = 1} \\ & = (\color{#3D99F6}{1})^2 - 2 \left(\color{#D61F06}{\frac {a+b+c}{abc}} \right) & \small \color{#D61F06}{\text{Note that } a+b+c = - abc \implies \frac {a+b+c}{abc} = -1} \\ & = 1 - 2(\color{#D61F06}{-1}) \\ & = \boxed{3} \end{aligned}$

$a+b+c=-abc$

Dividing $abc$ on both the sides.

$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac} = -1..........(1)$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$

Squaring on both sides,

$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=1$

$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{b^2}-2=1$ ...................from(1)

$\therefore\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\boxed{3}$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

Squaring both sides yields $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=1$ ....(eq 1)

Now it is given that $a+b+c=-abc$ This implies $\frac{1}{ab}=\frac{-c}{a+b+c} , \frac{1}{bc}=\frac{-a}{a+b+c}, \frac{1}{ca}=\frac{-b}{a+b+c}$

Putting these values in above equation (1) we get $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{-c}{a+b+c} + \frac{-a}{a+b+c} + \frac{-b}{a+b+c})$ =1

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-2\frac{a+b+c}{a+b+c}=1$

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} -2 =1$

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\boxed{3}$