I want to use the implicit function theorem for a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$. The set $M\subset \mathbb{R}^3$ is defined by the equation $f(x,y,z)=\text{function rule}=0$ for every $(x,y,z) \in M$.
I need to show that the equation $f(x,y,z)=0$ is locally solvable for one of the variables $x,y,z$ at every point of a set $M\subset \mathbb{R}^3$ i.e. if there exist a neighborhood $U\subset \mathbb{R}$ of $x$ and a continuous differentiable function $g: U\rightarrow \mathbb{R}^2$ with $g(x)=(y,z)$ and $f(x,g(x))=0$ for every $x\in U$
The best tool I have for this is the implicit function theorem:
But if I calculate $$\frac{\partial f}{\partial (y,z)} =\begin{pmatrix}\frac{\partial f}{\partial (y)} (x,y,z)&,&\frac{\partial f}{\partial (z)} (x,y,z)\end{pmatrix}$$
I get a $1\times 2$-Matrix of the form , so the Matrix cannot be invertable because $\frac{\partial f}{\partial (y,z)} $ is not a square matrix.
It would only make sense to me if $\frac{\partial f}{\partial (z)}(x,y,z)=0$ for every $(x,y,z)\in M$ but that is not the case.
Did I misunderstand the exercise and/or the meaning of the IFT?
$(*)$ The function $f$ is part of a homework exercise. I don't want to post the function rule here because my question is more general and I have some reservations about posting homework questions online. I hope my question is still clear.