We have an equilibrium given by
$$h_1(x,y)=f(x)+z-y=0,$$
$$h_2(x,y)=x-g(y)=0$$.
The implicit function theorem then says that (omitting the arguments):
$$\frac{\partial x}{\partial z}=\frac{-\det\left(
\begin{matrix}
\frac{\partial h_1}{\partial z} & \frac{\partial h_1}{\partial y} \\
\frac{\partial h_2}{\partial z} & \frac{\partial h_2}{\partial y}
\end{matrix}\right)}{\det\left(
\begin{matrix}
\frac{\partial h_1}{\partial x} & \frac{\partial h_1}{\partial y} \\
\frac{\partial h_2}{\partial x} & \frac{\partial h_2}{\partial y}
\end{matrix}\right)},\quad \frac{\partial y}{\partial z}=\frac{-\det\left(
\begin{matrix}
\frac{\partial h_1}{\partial x} & \frac{\partial h_1}{\partial z} \\
\frac{\partial h_2}{\partial x} & \frac{\partial h_2}{\partial z}
\end{matrix}\right)}{\det\left(
\begin{matrix}
\frac{\partial h_1}{\partial x} & \frac{\partial h_1}{\partial y} \\
\frac{\partial h_2}{\partial x} & \frac{\partial h_2}{\partial y}
\end{matrix}\right)}.$$
My source for this is Mathematical Methods and Models for Economists by Angel de la Fuente, although I don't have the book to hand now and can't remind myself of the intuition.
This implies
$$\frac{\partial x}{\partial z}=\frac{g'(y)}{1-f'(x) g'(y)}$$
$$\frac{\partial y}{\partial z}=\frac{1}{1-f'(x) g'(y)}.$$
For the implicit function theorem to hold and this solution to be valid, we need $1-f'(x) g'(y)\neq0$.
More generally, the way this works is as follows: you write down a system of equations whose roots characterize the equilibrium:
$$F_1(\mathbf{x};a)=0,F_2(\mathbf{x};a)=0,\ldots,F_n(\mathbf{x};a)=0$$
(where $a$ is the parameter of interest). From them, we construct the vector-valued function
$$\mathbf{F}(x)=[F_1(\mathbf{x};a),F_2(\mathbf{x};a),\ldots,F_n(\mathbf{x};a)]$$
which has the Jacobian matrix
$$\mathbf J = \frac{d\mathbf F}{d\mathbf x}
= \begin{bmatrix}
\dfrac{\partial F_1}{\partial x_1} & \cdots & \dfrac{\partial F_1}{\partial x_m}\\
\vdots & \ddots & \vdots\\
\dfrac{\partial F_n}{\partial x_1} & \cdots & \dfrac{\partial F_n}{\partial x_m} \end{bmatrix}.$$
To calculate the derivative of $x_i$ with respect to $a$, we construct the modified Jacobian in which we replace the $i^{\text{th}}$ column with derivative WRT $a$ instead of $x_i$. So, for $x_1$ this would look like
$$\mathbf{J}_{x_1}=\begin{bmatrix}
\dfrac{\partial F_1}{\partial a} & \dfrac{\partial F_1}{\partial x_2} & \cdots & \dfrac{\partial F_1}{\partial x_m}\\
\vdots & \vdots&\ddots & \vdots\\
\dfrac{\partial F_n}{\partial a}&\dfrac{\partial F_n}{\partial x_2} & \cdots & \dfrac{\partial F_n}{\partial x_m} \end{bmatrix}.$$
The derivative of interest is then calculated as $$\frac{\partial x_i}{\partial a}=\frac{-\det\mathbf{J}_{x_i}}{\det\mathbf{J}}.$$
We need $\det\mathbf{J}\neq0$ for the implicit function theorem to be valid.