I have a function $f(x) = \frac{x^2-1}{x^4+1}$
The x-coordintes of the maximum are found by the derivative.
$f'(x) =\frac{(x^4+1)(2x)-(x^2-1)(4x^3)}{x^4+1}$
Setting it equal to zero and solving for x gives:
$x^4-2x^2-1=0$ which solves to $x=\pm 1$
However, there is a minimum when $x=0$.
From my picture i only got the x-coordinates for the maximums, why didn't i get the x-coordinate for the minimum as well?
Because you divided both sides by $x$, excluding the solution $x = 0$.
$$f'(x) =\frac{(x^4+1)(2x)-(x^2-1)(4x^3)}{x^4+1}$$
$$f'(x) = 0 \iff (x^4+1)(2x)-(x^2-1)(4x^3) = 0$$
$$2x^5+2x-4x^5+4x^3 = 0 \iff -2x^5+4x^3+2x = 0$$
Here, $x = 0$ is clearly a solution. After this, you can simplify both sides and continue as you did. (Alternatively, you can factor $x$ from the start, which yields $x = 0$ as a solution.)