Suppose I have some function that $g$ that satisfies
$$g (..)=\frac{a}{f(x)(x+a)^2}=1$$
Using the implicit function theorem, I can see when an increase in $a$ results in an increase in $x$:
$$-\frac{\partial g/\partial a}{\partial g/\partial x}= - \frac{x-a}{a(a+x)f'(x)+2f(x)a}$$
Am I applying this correctly? And is the negative sign before $\frac{\partial g/\partial a}{\partial g/\partial x}$ necessary?
So when $a>x$, an increase in $a$ results in an increase in $x$, assuming that $a(a+x)f'(x)+2f(x)a$ is positve?
Note that under presumed smoothness conditions, we have
$$dg(x,a)=\frac{\partial g(x,a)}{\partial a}\,da+\frac{\partial g(x,a)}{\partial x}\,dx$$
On the surface $g(x,a)\equiv 1$, $dg(x,a)=0$. Hence,
$$\frac{dx}{da}=- \frac{\frac{\partial g(x,a)}{\partial a}}{\frac{\partial g(x,a)}{\partial x}}$$