Suppose I have the following function
$$f(x_1,x_2)=\frac{ax_1}{ax_1+bx_2}$$
Where $a,b>0$ and $a>>b$. Then I have that
$$\frac{\frac{\partial f}{\partial x_1}}{\frac{\partial f}{\partial x_2}}=-\frac{x_2}{x_1}$$
Which implies that given an increase in $x_2$, $x_1$ needs to increase by $\frac{x_2}{x_1}$ in order to keep $f(x_1,x_2)$ constant. Converseley, given an increase in $x_1$, $x_2$ needs to increase by $\frac{x_1}{x_2}$ to keep $f(x_1,x_2)$ constant.
This seems counterintuitive. If $a>>b$ then, surely $x_1$ needs to increase less to keep $f(x_1,x_2)$ constant relative to $x_2$, than $x_2$ would have to increase following an increase in $x_1$ to keep $f(x_1,x_2)$ constant? Why does this relation not depend on either $a$ or $b$?
Let $c = b/a$ and note that $$f(x_1, x_2) = \frac 1{1 + c\dfrac{x_2}{x_1}} \approx 1 - c\dfrac{x_2}{x_1}$$ when $c \ll \dfrac{x_1}{x_2}$. In this form, it is clearer that $x_1$ and $x_2$ are on approximately equal footing.
Note that an increase in $x_2$ by $\epsilon$ decreases $f$ by roughly $\dfrac {c\epsilon} {x_1}$. If $x_1$ is increased by $\epsilon$,
$$\frac{x_2}{x_1 + \epsilon} = \frac{x_2}{x_1}\left(\frac 1{1 + \dfrac{\epsilon}{x_1}}\right) \approx \frac{x_2}{x_1} - \frac{x_2\epsilon}{x_1^2}$$ Thus $f$ increases by roughly $\dfrac{x_2c\epsilon}{x_1^2} =\dfrac{x_2}{x_1}\dfrac{c\epsilon}{x_1}$, which is $-\dfrac{x_2}{x_1}$ times the change from increasing $x_2$ by the same amount, as expected.