Total derivative with non-independent variables. Suppose I have three variables that constrained by:
$$
x = yz,\tag{*}
\label{xyz}
$$
and I want to express total derivative w.r.t $x$ through a derivation w.r.t $y$ and $z$. The problem is when I try to calculate it via the standard method:
$$
\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d}x} +  \frac{\partial f}{\partial z}\frac{\mathrm{d} z}{\mathrm{d}x}, 
$$
where partial derivatives are actual partials, I cannot calculate factors with total derivatives because of infinite permutation using \eqref{xyz} ($y$ depends on $z$ and $x$ and so on).

What am I missing here? Should be easy. Sorry for this silly question.

Update
Let now the function $f$ is no longer defined independently (see Christian Blatter's answer below) but depends on $x$ only: $f = f(x)$. Is the following reasoning true?
On the one hand
$$
\mathrm{d}f = \frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{d}x = \frac{\mathrm{d}f}{\mathrm{d}x}(y\mathrm{d}z + z\mathrm{d}y).\label{1}\tag{1} 
$$
On the other hand
$$
\mathrm{d}f = \frac{\partial f}{\partial y}\mathrm{d}y + \frac{\partial f}{\partial z}\mathrm{d}z.\label{2}\tag{2}
$$
Comparing eqs.\eqref{1},\eqref{2}, we obtain
$$
\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}x} = \frac1y\frac{\partial f}{\partial z}\\
\frac{\mathrm{d}f}{\mathrm{d}x} = \frac1z\frac{\partial f}{\partial y}
\end{align}
$$
so eventually
$$
\frac{\mathrm{d}f}{\mathrm{d}x} = \frac12\Big(\frac1y\frac{\partial f}{\partial z} + \frac1z\frac{\partial f}{\partial y}\Big).
$$
 A: (After the update the following no longer applies.)
It seems that you are given a function $(y,z)\mapsto x(y,z):=y z$ and independently of this a function $(y,z)\mapsto f(y,z)$. Under these circumstances you want to know the "total derivative" ${df\over dx}$. 
Now for such a derivative to make sense it would be necessary that the $f$-value is uniquely determined by the $x$-value, i.e., that there is some sort of composition $\Phi(x)=f\bigl(\psi(x)\bigr)$. But this is not the case, since there are infinitely many points $(y,z)$ leading to the same $x$-value, and $f$ may take arbitrarily different values in these points.
Consider your example: Assume $f(y,z):=y$. Knowing the value $x:=yz$ does not determine the value of $y$,  hence of $f$.
Supplement after your update:
If you are given a function a function $x\mapsto f(x)$ of one variable $x$, and if this $x$ is made dependent on two other variables $y$, $z$ as follows: $(y,z)\mapsto x(y,z):=y z$, then we can define the composition
$$\Phi(y,z) :=f\bigl(x(y,z)\bigr)\ ,$$
a function of two variables. By the chain rule this function has partial derivatives
$$\Phi_y(y,z)=f'\bigl(x(y,z)\bigr)x_y(y,z)=f'\bigl(x(y,z)\bigr)z,\qquad
\Phi_z(y,z)=f'\bigl(x(y,z)\bigr)x_z(y,z)=f'\bigl(x(y,z)\bigr)y\ .$$
