I'm struggling in the resolution of this exercise. When he derives $f$, I don't understand the notation. What is $f$x1 supposed to mean? I think it means $\frac{\partial f1}{\partial x} $, being $f$1 the first component of the function. However, when he derives respect $y$ it keeps mantaining $f$x1 and that confuses me.
I've attached the resolution of the exercise in two pictures Part1 Part2 It's in Spanish, but here's the translation.
Consider the function $f: \mathbb R^2 \rightarrow \mathbb R $, $C^1$ . Knowing that the equation satisfies the implicit function theorem and it allows us to obtain $z(x, y)$ in the neighborhood of any given point. Calculate...
Resoultion Deriving respect $x$. Deriving respect $y$....
What your professor seems to mean (My mother tongue is English-I'm okay in French and Italian so I'm going by analogy) is $$u=x^2-y^2, v=y^2-z^2,f(u,v)=0.$$ $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=0,$$ $$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=0.$$ $$\frac{\partial u}{\partial x}=2x,\frac{\partial u}{\partial y}=-2y,$$ $$\frac{\partial v}{\partial x}=-2z\frac{\partial z}{\partial x},\frac{\partial v}{\partial y}=2y-2z\frac{\partial z}{\partial y}.$$