How to determine $x$? 
If $f\left(\dfrac{x-1}{x+1}\right)=x^2+x-1$. What is $x=?$
A) $0\quad$ B) $1\quad$ C)$-1\quad$ D)$2\quad$ E)$-2$

I believe, the problem is missing something, because for every $x$ I plug in, there exists a value of the function.

As has been pointed out in the comments, there is clearly something amiss in the problem as it is stated here. To underscore that fact, notice that the function $f$ is implicitly defined by the equation
$$ f\left(\dfrac{x-1}{x+1}\right)=x^2+x-1 $$
If we let $t=\dfrac{x-1}{x+1}$ we find that
$$ f(t)=-\frac{t^2-4t+1}{(t-1)^2} $$
which is define for all $t\ne1$. But there is no value of $x$ for which $t=1$.
So there is nothing about the definition of the function which implies any specific value of $x$.
