Vtyjkyuk

Let $F(x,y,z)=0$. So $x,y,z$ are defined implicitly in function of the other variable, i.e. $x=x(y,z)$, $y=y(x,z)$ and $z=z(x,y)$. Now $$dx=fracpartial xpartial ydy+fracpartial xpartial zdz=fracpartial xpartial yleft(fracpartial ypartial xdx+fracpartial ypartial zdzright)+fracpartial xpartial zdz$$
and thus $$left(fracpartial xpartial yfracpartial ypartial x-1right)dx+left(fracpartial xpartial z+fracpartial xpartial yfracpartial ypartial zright)dz=0.$$
Since $dx$ and $dy$ are linearly independent, we finally get

beginalign*
fracpartial xpartial yfracpartial ypartial x&=1 tag1\
fracpartial xpartial yfracpartial ypartial z&=-fracpartial xpartial ztag2
endalign*

Equation $(1)$ is natural, but equation $(2)$ should be $fracpartial xpartial yfracpartial ypartial z=fracpartial xpartial zfracpartial ypartial y=fracpartial xpartial z,$ no ? So what's the matter here ? I know it's correct, but what does it mean exactly such a contradiction ? Because at the end I get
$$fracpartial xpartial yfracpartial ypartial zfracpartial zpartial x=-fracpartial xpartial zfracpartial zpartial x=-1$$
instead of $1$ (what should be expected). What is the mystery behind ? :)

Here is a more symmetric explanation of this computation that hopefully makes the result less surprising. Since $F$ is constant on our surface, we have $$0=dF=F_xdx+F_ydy+F_zdz.$$
Now, how does this relate to partial derivative like $fracpartial xpartial y$? That partial derivative is just the coefficient of $dy$ when we write $dx$ as a linear combination of $dy$ and $dz$. Solving the equation above, we have $$dx=-fracF_yF_xdy-fracF_zF_xdx.$$ So, $fracpartial xpartial y=-fracF_yF_x$. Intuitively, this makes sense: if $z$ is held constant and we vary $y$ in one direction, we need to vary $x$ in the direction that makes $F$ move in the opposite direction, to keep $F$ equal to $0$. So since $F_x$ and $F_y$ represent the directions that $F$ varies when we change $x$ and $y$, the minus sign comes from needing the changes in $F$ from $x$ and $y$ to cancel out.

But now we see immediately where your surprising minus signs are coming from. If we compute $fracpartial xpartial yfracpartial ypartial z$ we would get $$left(-fracF_yF_xright)left(-fracF_zF_yright)=fracF_zF_x.$$ Since we lost our minus sign, this is the negative of $fracpartial xpartial z=-fracF_zF_x$. Using the same intuition as before, we are multiplying the change in $x$ needed to counteract a change in $y$ (and keep $F$ constant) by the change in $y$ needed to counteract a change in $z$. The two "counteracts" cancel each other out, and we end up with the change in $x$ needed to duplicate a change in $z$, rather than to counteract the change in $z$.

Ultimately, the lesson here is that derivatives are not fractions, especially not partial derivatives. Ordinary derivatives often behave like fractions (because they are limits of fractions) via the chain rule. Partial derivatives do not (and the chain rule for them does not look like just multiplying fractions!), because as explained in J.G.'s answer they are extremely sensitive to what other variables are being held constant. In particular, in your partial derivatives $fracpartial xpartial y$ and $fracpartial ypartial z$, $z$ is being held constant for the first one while $x$ is being held constant for the second one. So these derivatives are being computed under different "background assumptions", and it's not particularly reasonable to expect them to behave like fractions the way single-variable derivatives do.

Partial derivatives are only defined once you specify what's held constant. Partial derivatives' product only allows chain-rule style cancellations if they're defined with the same thing(s) assumed constant. For example, if $x,,y,,z$ were non-constant differentiable functions of $w$, you'd have $(fracpartial xpartial y)_w(fracpartial ypartial z)_w(fracpartial zpartial x)_w=+1$, where the subscript indicates the constant-$w$ condition. The result you're trying to understand is radically different; it's $(fracpartial xpartial y)_z(fracpartial ypartial z)_x(fracpartial zpartial x)_y=-1$, with a condition $f(x,,y,,z)=0$ existing and no fourth variable involved.

As a simple example of why the choice of what to hold constant matters, compare $2$-dimensional Cartesian and polar coordinates. Holding $y$ constant, $x^2=r^2-y^2$ implies $2x(fracpartial xpartial r)_y=2r$ so $(fracpartial xpartial r)_y=fracrx$; holding $theta$ constant, $x=rcostheta$ implies $(fracpartial xpartial r)_theta=costheta=fracxr$.