how to solve

y=0 basically means the x axis.

if it wasnt for the$f(x)+g(x)$ condition, you could just chuck a negative sign in front of it and you were good to go.

anyway -- we are given that   $f(x) = \frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) }$ -- now what is the value of the polynomial g(x) so that it satisfies the condition? lets set up this 'equation' : 

 $\frac{\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}+g(x)=\frac{-\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}$  

 $g(x)=-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4}-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4}$ 

 $g(x)=-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4}$ 

lets confirm the condition:

 $f(x)+g(x)$ $\Rightarrow$ $\frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) } +-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4}  = \boxed{\frac{-3x^4-3x^3-3x^2-3}{x^2+2x-4}}$ 

if you graph it you would get:

: 

here is the link for it as well: https://www.desmos.com/calculator/v2w0ukjkoq

:D

As follows: