I have been learning about implicit differentiation and so far I have not much trouble when I am given an equation where I can see the variables explicitly (like $x^2+y^2-1 = 0$). However, I have stumbled upon a problem that has me stuck for too long now and would like some guidance. Here it goes
Suppose $f: U \subset \mathbb{R}^2 \to \mathbb{R}$ is a differentiable function defined in the open set $U$ and $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ are continuous at the point $(-1,2)$. Furthermore suppose that $f(-1,2) = 5$ and $\frac{\partial f(-1,2)}{\partial x}\frac{\partial f(-1,2)}{\partial x} > 0$. I am asked the following:
a) To show that $f(x,y) = 5$ defines implicitly the functions $x=\phi(y)$ and $y=\psi(x)$ in the vicinity of $(-1,2)$
b) To calculate $\phi'(-1)\psi'(2)$.
Setting $F(x,y) = f(x,y) -5$, I was able to verify part a) using the implicit function theorem. However, I am stuck at b). Normally I would go on by calculating the total differential of $F$, that is
$$ \begin{align} \mathrm{d}F &= \frac{\partial f}{\partial x}\mathrm{d}x + \frac{\partial f}{\partial y}\mathrm{d}y = 0 \\ \mathrm{d}F &= \frac{\partial f}{\partial \psi}\frac{\mathrm{d} \psi}{\mathrm{d} x}\mathrm{d}x + \frac{\partial f}{\partial \phi}\frac{\mathrm{d} \phi}{\mathrm{d} x}\mathrm{d}y = 0 \end{align} $$
But that's it I am afraid. I don't know how to use the information provided to say something about the partial derivatives of $f$ i.e., I would normally be able to calculate them and solve for the desired derivative, but I can't proceed that way. Any help would be greatly appreciated.
Note $$x=\phi(y)\Rightarrow 1=\frac{dx}{dx}=\frac{d}{dx}\phi(y)=\phi'(y)\frac{dy}{dx}$$ and $$y=\psi(x)\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\psi(x)=\psi'(x)$$ From these two equations you obtain $$1=\phi'(y)\frac{dy}{dx}=\phi'(y)\psi'(x)$$ So for $(x,y)$ in a neighborhood of $(-1,2)$ we have $\phi'(y)\psi'(x)=1$. In particular $\phi'(2)\psi'(-1)=1$.