Above is implicit function theorem, and here is a special case
In second case, $f(y)=(x,z)\in\mathbb{R}^2$, also $F\in\mathbb{R}^2$, how do we find $\partial_y f$ ? I know $\partial_y f=-\frac{\partial_y F}{\partial_y F}$,
but this only apply if $F\in\mathbb{R}$ right?
In the most general case of the Implicit Function Theorem we have that if $F\colon\mathbb{R}^{n+m}\to\mathbb{R}^m$ and $f\colon\mathbb{R}^n\to\mathbb{R}^m$ is such that $F(x_1,\ldots,x_n,f(x_1,\ldots\,x_n))=0$, then $$\begin{pmatrix}\frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n}\\\vdots & \ddots & \vdots\\\frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n}\end{pmatrix} = -\begin{pmatrix}\frac{\partial F_1}{\partial y_1} & \cdots & \frac{\partial F_1}{\partial y_m}\\\vdots & \ddots & \vdots\\\frac{\partial F_m}{\partial y_1} & \cdots & \frac{\partial F_m}{\partial y_m}\end{pmatrix}^{-1}\cdot\begin{pmatrix}\frac{\partial F_1}{\partial x_1} & \cdots & \frac{\partial F_1}{\partial x_n}\\\vdots & \ddots & \vdots\\\frac{\partial F_m}{\partial x_1} & \cdots & \frac{\partial F_m}{\partial x_n}\end{pmatrix}$$
Therefore in this particular example for the functions $F\colon\mathbb{R^3}\to\mathbb{R}^2$ and $f\colon\mathbb{R}\to\mathbb{R}^2$ we have $$D_yf=\begin{pmatrix}\frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial y}\end{pmatrix} = -\begin{pmatrix}\frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial z}\\\frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial z}\end{pmatrix}^{-1}\cdot\begin{pmatrix}\frac{\partial F_1}{\partial y}\\\frac{\partial F_2}{\partial y}\end{pmatrix}$$