well let's see... If I'm understanding you, there should be an = 0 at the end of that expression making this a quadratic (it might not be at all though).
so let's say 3/x^2 - 1 = 0
Firstly, you don't want a denominator in the equation, it makes it much harder to solve. So to move it out of the bottom, multiply the entire equation by x^2 : (3/x^2 - 1 = 0)*(x^2)
The denominator of 3/x^2 will be canceled out because it is being multiplied by its inverse (2 and 1/2 are inverses. x^2 and 1/x^2 are inverses) leaving a 3 in the numerator (3/1 or just 3).
The -1 term also gets multiplied by x^2 and yields -x^2
And 0 also gets multiplied and anything times 0 equals 0.
So you now have 3 - x^2 = 0 which can be rewritten as -x^2 + 3 = 0
A personal preference is to multiply the entire equal by -1 to make the x^2 term positive (so now x^2 - 3 = 0)
This can be forced to be the difference of two squares (its a special type of quadratic that always breaks down into two conjugates - binomials that multiply to cancel their like terms)
The equation would factor to (x + sqr(3))(x - sqr(3)) = 0 (its not a perfect difference of two squares but it works because sqr(3)^2 = 3. I cannot better explain it. Sorry if that makes you confused).
But now, since you now have two x terms, you solves each for 0 making (x + sqr(3) = 0) and (x - sqr(3) = 0)
Solve each by moving the sqr(3) term to the other side and you get x = +or- sqr(3)
This is the answer to your problem IF IT IS WHAT I THINK IT IS! Next time, please try to be more specific
