So I stumbled into this problem, and I honestly have no idea what rules to use. I'd appreciate some help, and if you could direct me to some useful resource to learn the right method I'd be thankful.
Here's the question :
$y = f(x,p) \quad ; x = g(p,y)$
Calculate the derivative of y with respect to x in terms of the 4 partial derivatives we suppose known :
$\dfrac{\partial f}{\partial x} \quad ; \quad \dfrac{\partial f}{\partial p} \quad ; \quad \dfrac{\partial g}{\partial p} \quad ; \quad \dfrac{\partial g}{\partial y}$
$$y = f(x,p) \quad ;\quad dy=\dfrac{\partial f}{\partial x} dx+ \dfrac{\partial f}{\partial p}dp$$
$$x = g(p,y) \quad ;\quad dx=\dfrac{\partial g}{\partial p} dp+ \dfrac{\partial g}{\partial y}dy$$
$$dp=\frac{dy-\dfrac{\partial f}{\partial x} dx}{\dfrac{\partial f}{\partial p}} = \frac{ dx- \dfrac{\partial g}{\partial y}dy}{\dfrac{\partial g}{\partial p}}$$
$$\dfrac{\partial g}{\partial p}\left(dy-\dfrac{\partial f}{\partial x} dx\right) = \dfrac{\partial f}{\partial p}\left(dx-\frac{ \partial g}{\partial y}dy\right)$$
$$\dfrac{\partial g}{\partial p}\left(\frac{dy}{dx}-\dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial p}\left(1-\frac{ \partial g}{\partial y}\frac{dy}{dx}\right)$$
$$\frac{dy}{dx}\left(\dfrac{\partial g}{\partial p} + \dfrac{\partial f}{\partial p}\frac{ \partial g}{\partial y} \right)= \dfrac{\partial g}{\partial p}\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial p} $$
$$\frac{dy}{dx} = \frac{\dfrac{\partial g}{\partial p}\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial p}}{\dfrac{\partial g}{\partial p} + \dfrac{\partial f}{\partial p}\dfrac{ \partial g}{\partial y} }$$