$a,b\in\mathbb{R}\ni a$a,b\in\mathbb{R}\ni a,b>0, a<b<{1\over a}$ and $$x=(a+{1\over a})-(b+{1\over b})$$
Then $a) x>0$ 
b) $x<0$
$c) x=0$
d) no such conclusion can be drawn  about $x$
just confirm me that answer is $d$?

Notice that $a < 1$, because $a<\frac1a$. If $a < b < \frac 1 a$, then also $a < \frac 1 b < \frac 1 a$, so without loss of generality $b \leq 1$ and we have
$$x = (a+\frac1a) - (b + \frac1b) = a-b + \frac1a-\frac1b =a-b - \frac{a-b}{ab} =\\= \frac{ab(a-b) - (a-b)}{ab} = \frac{(a-b)(ab-1)}{ab} = \frac{(b-a)(1-ab)}{ab}$$
Now $b-a > 0$, $ab>0$, and also $1-ab > 1-b \geq 0$, so in the end
$$x = \frac{(b-a)(1-ab)}{ab} > 0$$
