Suppose a space curve $\gamma$ is given via
$$\gamma:\begin{cases}\Phi_1(x,y,z)=0,\\\Phi_2(x,y,z)=0\end{cases}$$ where $x,y,z$ are, in turn, functions of $t$:
\begin{cases}x=\phi(t),\\y=\psi(t),\\z=\chi(t).\end{cases}
The task is to find tangent line equation to the curve $\gamma.$ So far, I have shown that
$$\frac{\frac{\mathrm{dx}}{\mathrm{dt}}}{\frac{\partial\Phi_1}{\partial y}\frac{\partial\Phi_2}{\partial z}-\frac{\partial\Phi_1}{\partial z}\frac{\partial\Phi_2}{\partial y}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\partial\Phi_1}{\partial z}\frac{\partial\Phi_2}{\partial x}-\frac{\partial\Phi_1}{\partial x}\frac{\partial\Phi_2}{\partial z}}=\frac{\frac{\mathrm{dz}}{\mathrm{dt}}}{\frac{\partial\Phi_1}{\partial x}\frac{\partial\Phi_2}{\partial y}-\frac{\partial\Phi_1}{\partial y}\frac{\partial\Phi_2}{\partial z}}.\ \ \ \ (1)$$
The form given in the textbook is
$$\frac{X-x}{\begin{vmatrix}\frac{\partial\Phi_1}{\partial y} & \frac{\partial\Phi_1}{\partial z} \\ \frac{\partial\Phi_2}{\partial y} & \frac{\partial\Phi_2}{\partial z}\end{vmatrix}}=\frac{Y-y}{\begin{vmatrix}\frac{\partial\Phi_1}{\partial z} & \frac{\partial\Phi_1}{\partial x} \\ \frac{\partial\Phi_2}{\partial z} & \frac{\partial\Phi_2}{\partial x}\end{vmatrix}}=\frac{Z-z}{\begin{vmatrix}\frac{\partial\Phi_1}{\partial x} & \frac{\partial\Phi_1}{\partial y} \\ \frac{\partial\Phi_2}{\partial x} & \frac{\partial\Phi_2}{\partial y}\end{vmatrix}},\ \ \ (2)$$
or, equivalently,
$$\frac{X-x}{\frac{\partial\Phi_1}{\partial y}\frac{\partial\Phi_2}{\partial z}-\frac{\partial\Phi_1}{\partial z}\frac{\partial\Phi_2}{\partial y}}=\frac{Y-y}{\frac{\partial\Phi_1}{\partial z}\frac{\partial\Phi_2}{\partial x}-\frac{\partial\Phi_1}{\partial x}\frac{\partial\Phi_2}{\partial z}}=\frac{Z-z}{\frac{\partial\Phi_1}{\partial x}\frac{\partial\Phi_2}{\partial y}-\frac{\partial\Phi_1}{\partial y}\frac{\partial\Phi_2}{\partial z}}.\ \ \ (2')$$ Here $(X,Y,Z)$ are the coordinates of any point on the tangent line.
Question: What is the transition from $(1)$ to $(2')$?
My answer to your question is a rewriting of your proportionalities under the following vectorial terms:
$$\tag{1}\vec{S}=\pmatrix{\frac{\mathrm{dx}}{\mathrm{dt}}\\ \frac{\mathrm{dy}}{\mathrm{dt}} \\ \frac{\mathrm{dz}}{\mathrm{dt}}} \ \ \text{prop}^{al} \ to \ \ \vec{N_1} \times \vec{N_2}=\pmatrix{\frac{\partial\Phi_1}{\partial x}\\ \frac{\partial\Phi_1}{\partial y} \\ \frac{\partial\Phi_1}{\partial z}} \times \pmatrix{\frac{\partial\Phi_2}{\partial x}\\ \frac{\partial\Phi_2}{\partial y} \\ \frac{\partial\Phi_2}{\partial z}} \ \ \text{prop}^{al} \ to \ \vec{M_0M}=\pmatrix{X-x_0\\Y-y_0\\Z-z_0}$$
(where $\vec{N_k}$ is a normal vector in point $M_0 (x_0,y_0,z_0)$ to the surface with equation $\Phi_k(x,y,z)=0$ for $k=1,2$).
In fact, (1) describes plainly 3 ways to express a directing vector of the tangent line at $M_0.$