I would appreciate some help to understand in a rigorous way a physics development.
I know the analytical expression of two quantities $S(V,T,\mu)$ and $N(V,T,\mu)$, and need to calculate $\left(\frac{\partial S}{\partial N}\right)_{V,\,T}$.
From chain rule, I know I should consider $$\left(\frac{\partial S}{\partial N}\right)_{V,\,T} = \left(\frac{\partial S}{\partial \mu}\right)_{V,\,T} \left(\frac{\partial \mu}{\partial N}\right)_{V,\,T}$$
but I'm not sure how to estimate the second term. I see no reasons why $\left(\frac{\partial \mu}{\partial N}\right)_{V,\,T}=\left(\frac{\partial N}{\partial \mu }\right)^{-1}_{V,\,T}$, so the only option I see is to estimate $\mu$ as a function of $N$, and then differentiate the relation. But there is no analytical expression for it, so I guess numerics is the only way ? Or am I missing something ?
Imagine you have an implicit relation between $x$ and $y$: $$f(x,y) = 0$$ Then the total differential of $f$ is exactly $0$. For your case let $y$ depend on $x$, clearly: $$\frac{df(x,y(x))}{dx}=\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}=0$$ As a result, we can obtain the derivative of $y$ wrt. $x$ as follows: $$\frac{dy}{dx} = -\frac{\partial f}{\partial x}\bigg/\frac{\partial f}{\partial y} \tag{*}$$
The point is that you know how to compute analytically or numerically the derivatives involved in $(*)$
You have to compute $\frac{\partial S}{\partial N}$ and you know by the chain rule that: $$\frac{\partial S(V,T,\mu(V,T,N))}{\partial N}=\frac{\partial S}{\partial \mu}\frac{\partial \mu}{\partial N}$$ Since you do not have an explicit relation like $\mu = \mu(V,T,N)$ you have to apply $(*)$ to the function $f(T,V,\mu,N)=N-N(T,V,\mu)=0$ with $x=N$ and $y=\mu$. Therefore you will have that the sought derivative is simply: $$ \frac{\partial \mu}{\partial N}=-\frac{\partial f}{\partial N}\bigg/\frac{\partial f}{\partial \mu}=-1\bigg/\frac{\partial N}{\partial\mu}$$