1. You mention the method to be used for the solution, Vieta.
If the roots of the cubic are a, b and c, then
(x−a)(x−b)(x−c)=0,expanding,x3−x2(a+b+c)+x(ab+bc+ca)−abc=0.
So, for the given cubic,
a+b+c=3……(1)ab+bc+ca=−15/2……(2)abc=3/2……(3).
It's just follow your nose after that.
1a2+1b2+1c2=a2b2+b2c2+c2a2a2b2c2.
The denominator is easy, just substitute (3).
For the numerator it looks like you have to make use of (2), so,
(ab+bc+ca)2=a2b2+b2c2+c2a2+2a2bc+2ab2c+2abc2=a2b2+b2c2+c2a2+2abc(a+b+c),
and now it's just substitution from (1), (2) and (3).