Implicit partial derivatives Say I have a function $F(P,V,T) = 0 $.
Using the chain rule: $$dF = \frac{\partial F}{\partial P}dP + \frac{\partial F}{\partial V}dV + \frac{\partial F}{\partial T}dT$$ Now, if P is constant, we have $dP = 0$ and $dF = 0$ of course, because $F$ is a constant (0). We get:
$$dF = \frac{\partial F}{\partial V}dV + \frac{\partial F}{\partial T}dT = 0 \implies (\frac{dV}{dT})_P = - \frac{\frac{\partial F}{\partial T}} {\frac{\partial F}{\partial V}}$$ And that is true generally? Correct me if i'm mistaken please. Now, if we have an explicit expression for $P(T,V)$ we can also say for constant P: $$ (\frac{dV}{dT})_P = - \frac{\frac{\partial P}{\partial T}} {\frac{\partial P}{\partial V}}$$
BUT the results aren't always the same, why is that? Do P,V and T have to be independent of each other for the first derivation to be correct? I'm really confused and will be glad for explanation without getting too rigorous.
Edit: Example (from physics):
Say:$$F(P,V,T)=PV-PNb+\frac{aN^{2}}{V}-\frac{aN^{3}b}{V^{2}}-Nk_{B}T=0$$
We can solve for P:
$$P = N\left(\frac{K_{B}T}{V-Nb}-\frac{aN}{V^{2}}\right)$$
We now get:
$$- \frac{\frac{\partial F}{\partial T}} {\frac{\partial F}{\partial V}} \neq - \frac{\frac{\partial P}{\partial T}} {\frac{\partial P}{\partial V}} $$
 A: Short Answer
They are the same, including in your example. You either did the algebra wrong or forgot one key step at the end.
Long Answer
Yes, your thoughts are correct. If you can write $P$ as a function, then for constant $P$ satisfying $F(P,V,T)=0$, we do have $\dfrac{\mathrm dV}{\mathrm dT}=-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}=-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$.
Example Case
In your particular example, $F(P,V,T)=PV-PNb+\frac{aN^{2}}{V}-\frac{aN^{3}b}{V^{2}}-Nk_{B}T=0$ we can indeed solve for $P$ to find $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$.
Then we can calculate the two fractions. For $-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}$, we get something like $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$. But for $-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$ we get something like $\dfrac{k_BNV^3}{aN^2(2Nb-V)+PV^3}$, which looks different. But we have to remember to substitute in the constant value of $P$ from $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$. Then we get $\dfrac{k_BNV^3}{aN^2(2Nb-V)+N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)V^3}$, which simplifies down to the same $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$.
You can verify the calculation with Wolfram/Mathematica code:
Wolfram Language (Mathematica), 458 bytes
Print[Simplify[Solve[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T == 0, P]]]
Print[Simplify[-D[N (-((kB T)/(b N - V)) - (a N)/V^2), T]/D[N (-((kB T)/(b N - V)) - (a N)/V^2), V]]]
Print[Simplify[-D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, T]/D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, V]]]
Print["Why Not the same? Substitute in for P."]
Print[Simplify[(kB N V^3)/(a N^2 (2 b N - V) + P V^3) /. P -> N (-((kB T)/(b N - V)) - (a N)/V^2)]]

Try it online!
