There is not a simple geometric derivation of this fact. I will slightly reshape the discussion I wrote in the posts I already linked to you. The key idea is the implicit differentiation formula that introduces a negative sign: If $f(x,y)=c$ defines $y$ locally as a function $\phi$ of $x$, then
$$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}.$$
You can think of this geometrically in terms of the tangent line to $y=\phi(x)$ at $(a,b)$. Working with linear approximation along that graph, we have
$$0=\Delta f = \frac{\partial f}{\partial x}(a,b)\Delta x + \frac{\partial f}{\partial y}(a,b)\Delta y,$$ and so, indeed,
$$\phi'(a) \approx \frac{\Delta y}{\Delta x} = -\frac{\frac{\partial f}{\partial x}(a,b)}{\frac{\partial f}{\partial y}(a,b)}.$$
Now, consider a level surface of a function of three variables. We start with $f(x,y,z) = c$ and assume this (locally) defines each of the variables as a differentiable function of the remaining two. It will now be important for clarity to introduce the notation — common in thermodynamics — of indicating with a subscript what variable is held fixed when we compute a partial derivative. From the two-variable formula I derived above, we get
\begin{align*}
\left(\frac{\partial z}{\partial x}\right)_y &= -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}, \\
\left(\frac{\partial x}{\partial y}\right)_z &= -\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}, \\
\left(\frac{\partial y}{\partial z}\right)_x &= -\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}.
\end{align*}
Multiplying the three together, we get the result you want. Not very geometric, but it's all about implicit differentiation and tangent lines (planes).
I will remark that with partial derivatives it is generally false that
$\dfrac{\partial z}{\partial x} = \dfrac 1{\dfrac{\partial x}{\partial z}}$. For example: Consider cartesian and polar coordinates, with
$$x=r\cos\theta,\ y=r\sin\theta \qquad\text{and}\qquad r=\sqrt{x^2+y^2}, \ \theta = \arctan(y/x).$$
Then $\dfrac{\partial x}{\partial\theta} = -r\sin\theta = -y$, whereas $\dfrac{\partial\theta}{\partial x} = -\dfrac y{x^2+y^2}$. However, it is true, as you can check, that $\left(\dfrac{\partial\theta}{\partial x}\right)_y = \dfrac 1{\left(\dfrac{\partial x}{\partial\theta}\right)_y}$.
