If for some function $f$ (assumed to be twice differentiable), we have that
$$\frac{df}{dx_1}(x_1)>\frac{df}{dx_2}(x_2) \iff x_1<x_2$$
What needs to be true about its second derivative? Since the slope of $f$ is decreasing in its argument, it seems like its second derivative should be negative (concave). However, I have an example that satisfies the above inequality but is convex. Is this possible?
My example is
$$\frac{\partial f(x)}{\partial x}=-\frac{a}{(a*k)+2x}$$
Where $f(x)$ is defined implictly, and $a,k>0$. Here,
$$-\frac{a}{(a*k)+2x_1}>-\frac{a}{(a*k)+2x_2} \iff x_1 < x_2 $$
However, $\frac{\partial^2f(x)}{\partial^2x}>0$ for every $x$.
I guess this inequality can hold for a function that is either strictly concave with a positive derivative, or strictly convex and with a negative derivative. My example is is the second type.
Your example does not have the properties you claim it has. If we insert $a=b=1$, $x_1=1$, $x_2=2$, we get $$-\frac{1}{1\cdot 1+2\cdot 1}>-\frac{1}{(1\cdot 1)+2\cdot 2} \iff 1 < 2$$ The right-hand side of this is true (because $1$ is indeed less than $2$) but the lefth-hand side is false: $-\frac13$ is not larger than $-\frac15$.
Indeed, the function $x\mapsto -\frac{a}{ak+2x}$ with $a,k>0$ is strictly increasing on each of the connected components of its domain -- that is, it is increasing on $(-\infty,-\frac12ak)$ and increasing on $(-\frac12ak,\infty)$.
(Putting the $x$ in denominator flips the inequality once, negating the result flips it once again).
This agrees with its derivative being positive everywhere.