In this extra explanation we show that the slope ($\text{s}$) of the tangent line to the level curve of the function $z(x,y)$ with value $k$ is equal to
$$\text{s} = -\dfrac{z'_x(x,y)}{z'_y(x,y)}.$$
The general form of the tangent line is $ax + b$, where $a$ is the slope. The value of $a$ is determined by the derivative of $y(x)$, where $y(x)$ is the function such that the tangent line is tangent to its graph. Since we deal with a level curve there is no function $y(x)$ given, but this function is determined by
$$ z(x,y) = k.$$
Hence, there exists a function $y(x)$ such that $z(x,y(x)) = k$. We differentiate both sides of this equation with respect to $x$, where we use the Chain rule: special case 2:
$$
\begin{align}
z(x,(y(x)) &= k\\
&\hspace{-1cm}\text{Derivative with respect to~}x\\
z'_x(x,y(x)) + z'_y(x,y(x))\cdot y'(x) &= 0\\
&\hspace{-1cm}\text{Solving for~}y'(x)\\
z'_y(x,y(x))y'(x) &= - z'_x(x,y(x))\\
y'(x) &= -\dfrac{z'_x(x,y(x))}{z'_y(x,y(x))}.
\end{align}
$$
The final step is of course only allowed when $z'_y(x,y) \neq 0$. Finally, we know that $y(x) = y$, hence
$$\text{s} = y'(x) = -\dfrac{z'_x(x,y)}{z'_y(x,y)}.$$
