Do you mean: (1/x)+x=3, what does 1/x^2+x^2=?
Multiply first equation through by x: 1+x^2=3x. 1/x^2+x^2=(x^4+1)/x^2=(x^4+1)/(3x-1)=$\dfrac{x^4+1}{3x-1}$.
Solving the quadratic: x^2-3x+1=0 we get x=3/2+sqrt(5)/2=2.618 or 0.382 approx., making x^2=6.8541 or 0.1459. 1+x^2=3x=7.8541 or 1.1459.
(x^4+1)/(3x-1)=7, because x^2=3x-1=3(3/2+sqrt(5)/2)-1=7/2+3sqrt(5)/2. x^4=49/4+21sqrt(5)+45/4=47/2+21sqrt(5)/2. Therefore, x^4+1=49/2+21sqrt(5)/2 and 7(3x-1)=49/2+21sqrt(5)/2.
The solution is $\dfrac{x^4+1}{3x-1}$=7.