When $f$ is a function of two variables and writing $z=f(x,y)$, setting $z=f(x,y)=\mathrm{constante}$ defines an implicit function $y=g(x)$ and we have the relation
$$\frac{dy}{dx}=-\frac{∂f/∂x}{∂f/∂y} \space \space \space \mathrm{or} \space \space \space \frac{dy}{dx}=-\frac{∂z/∂x}{∂z/∂y}.\tag{1}\label{Eq1}$$
On the other hand, if $f(x,y,z)=0$ defines a surface, the equation defines the implicit functions
$$x=x(y,z),$$ $$y=y(x,z),$$ $$z=z(x,y),$$
which satisfy the relation
$$\left(\frac{∂x}{∂y}\right)_{z}\left(\frac{∂y}{∂z}\right)_{x}\left(\frac{∂z}{∂x}\right)_{y}=-1.\tag{2}\label{Eq2}$$
By rearranging the latter, we obtain
$$\left(\frac{∂x}{∂y}\right)_{z}=-\left(\frac{∂y}{∂z}\right)_{x}^{-1}\left(\frac{∂z}{∂x}\right)_{y}^{-1}$$
and using the fact that
$$\left(\frac{∂x}{∂y}\right)_{z}=1/\left(\frac{∂y}{∂x}\right)_{z}\space\space\mathrm{and}\space\space\left(\frac{∂z}{∂y}\right)_{x}=1/\left(\frac{∂y}{∂z}\right)_{x}$$
we obtain
$$\left(\frac{∂y}{∂x}\right)_{z}=-\frac{(∂z/∂x)_{y}}{(∂z/∂y)_{x}}$$
which looks a lot like the first relation $(1)$.
Questions
Are these two relations fundamentally the same?
If so, are the differences in notations ($d$ vs $∂$) just irrelevant (or maybe the $dy/dx$ in relation $(1)$ should rather be written $∂y/∂x$) ?
If not, how are they different?
Yes, the two relations describe the same idea. In fact, $(2)$ – the usual "Euler cyclic product" you find in thermodynamics textbooks – can be derived from (another instance of) $(1)$, which is a property of the derivative of an implicitly defined function. Consider a sufficiently regular (for example, smooth, but one can work with less) constraint $f(x,y,z)=0$ and a point $(x_0,y_0,z_0)$ where none of the partials of $f$ vanish. Then the implicit function theorem tells you that there are three (smooth) functions $X,Y,Z$ of two real variables, $$(y,z) \mapsto X(y,z), \qquad (x,z) \mapsto Y(x,z), \qquad (x,y)\mapsto Z(x,y), $$ such that $$f(X(y,z),y,z)=0, \qquad f(x,Y(x,z),z)=0, \qquad f(x,y,Z(x,y))=0,$$ for points sufficiently close to $(x_0,y_0,z_0)$. (The functions $X,Y,Z$ are like $g$ in $(1)$, but each depends on the two other variables, not one.) Moreover, these satisfy $$ \begin{cases} \displaystyle\frac{\partial X}{\partial y}(y_0,z_0) = - \frac{\frac{\partial f}{\partial y} (x_0,y_0,z_0)}{\frac{\partial f}{\partial x}(x_0,y_0,z_0)}, \\[1em] \displaystyle\frac{\partial Y}{\partial z}(x_0,z_0) = - \frac{\frac{\partial f}{\partial z} (x_0,y_0,z_0)}{\frac{\partial f}{\partial y}(x_0,y_0,z_0)},\\[1em] \displaystyle\frac{\partial Z}{\partial x}(x_0,y_0) = - \frac{\frac{\partial f}{\partial x} (x_0,y_0,z_0)}{\frac{\partial f}{\partial z}(x_0,y_0,z_0)}, \end{cases} $$ as can be ascertained by differentiating the constraint with respect to $x,y,z$ alternatively. Multiplying these three and reverting to the thermodynamical short-hand notation, you obtain formula $(2)$.
The reason one uses $d$ instead of $\partial$ in $(1)$ is that, in that case, the implicit function $y=g(x)$ is a real function of one real parameter $x$, and so the derivative can only be taken with respect to that parameter – the partial derivative $\partial/\partial x$ coincides with the total derivative $d/dx$.
Actually, the above derivation gives some insight into possible generalizations of the cyclic product formula. For one, the presence of the factor $-1$ is entirely due to the fact that we are working with an odd number of variables $(x,y,z)$ – or more precisely, that we are multiplying an odd number of implicit derivatives. Furthermore, one could certainly consider the "anticyclic" product of implicit derivatives $\partial X/\partial z$, $\partial Y/\partial x$, $\partial Z/\partial y$, and obtain an analogous relation. More interestingly, if one works with $n$ variables connected by a scalar constraint like $f$, one has $!n$-many generalizations of the triple product formula (with $(-1)^n$ on the RHS), where $!n$ is the number of permutations of $n$ elements with no fixed point (or derangements) – including non-cyclic permutations, which are not an option when $n=3$. For example, when $n=4$, we have a quadruple product formula: $$\left(\frac{\partial x}{\partial y}\right)_{z,w} \left(\frac{\partial y}{\partial x}\right)_{z,w} \left(\frac{\partial z}{\partial w}\right)_{x,y} \left(\frac{\partial w}{\partial z}\right)_{x,y} = 1.$$ Funnily, there is nothing inherently cyclic about the cyclic product formula, if not in the very peculiar case $n=3$.