Differentiation with dependent variable Let 
$$
F(x, y) = x^3 + 7 y^2 x^4 - (2 x - y)^3
$$
and let $y=f(x)=x^2+1$.
Is it correct to write
$$
\frac{\partial F}{\partial y}=\frac{\partial F}{\partial x}\frac{\partial x}{\partial y}?
$$
If yes then why 
$$
\left.\frac{\partial F}{\partial y}\right|_{y=f(x)}\neq\left.\frac{\partial F}{\partial x}\frac{\partial x}{\partial y}\right|_{y=f(x)}
$$
This is what I tried 
F[x_, y_] := x^3 + 7 y^2 x^4 - (2 x - y)^3

TT := x^2 + 1
X1 := F[x, y] // D[#, y] & // ReplaceAll[#, {y -> TT}] & // ExpandAll
X2 := F[x, y] // D[#, x] & // Times[#, 1/D[TT, x]] & // 
   ReplaceAll[#, {y -> TT}] & // ExpandAll

Why $X1\neq X2$? 
 A: No, that is not correct.
There is ambiguity here because $z = F(x,y) = F(x,f(x))$ can be considered to depend on $x$ in two different ways. 
In order to clarify the situation, I'm going to introduce a third variable $t$  (which will in fact have the same value as $x$ but be considered a distinct variable). Then we can say that $z$ depends on the pair $(x,y)$ via the equation $z = F(x,y)$, and the pair $(x,y)$ in turn depends on $t$ via the equations $x = t$, $y = f(t)$. In other words, $z = F(t,f(t))$. We consider $z$ to be the highest level variable, $x, y$ to be at the middle level, and $t$ to be at the lowest level.
Now it is a straightforward matter to compute the required derivatives. We have
$$\frac{\partial z}{\partial x} = \frac{\partial F}{\partial x} = 3x^2 + 28x^3 y^2 - 6(2x - y)^2,$$
and
$$
\begin{align*}
\frac{dz}{dt} &= \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \\
&= \frac{\partial F}{\partial x} \cdot 1 + \frac{\partial F}{\partial y} \frac{df}{dt} \\
&= 3x^2 + 28x^3 y^2 - 6(2x - y)^2 + [14x^4 y + 3(2x-y)^2]\cdot 2t\\
&= 3t^2 + 28t^3 (t^2 + 1)^2 - 6[2t-(t^2 + 1)]^2 + (14t^4 (t^2 + 1) + 3[2t-(t^2 + 1)]^2)\cdot 2t
\end{align*}$$
We note that this result is different than the result we obtained for $\frac{\partial z}{\partial x}$ above (even when $x$ and $y$ are replaced with their values in terms of $t$). However, it is the same as the result we obtain for $\frac{dz}{dt}$ working directly from the expression
$$z = F(t,f(t)) = t^3 + 7t^4(t^2 + 1)^2 - [2t - (t^2 + 1)]^3.$$
Thus we see that the ambiguity was introduced into the notation by using the same variable $x$ at two different levels, even though the value of the variable is the same at both levels. 
The meaning of the notation $\frac{\partial z}{\partial x}$ in our example with $t$ is "the rate of change of $z$ with respect to $x$ when $y$ is held constant". The meaning of $\frac{dz}{dt}$ is "the rate of change of $z$ with respect to $t$, where $x$ and $y$ are permitted to vary as functions of $t$."
We see that $\frac{\partial z}{\partial x}$ and $\frac{dz}{dt}$ have completely different meanings. The question that arises is how we should interpret the notations $\frac{\partial z}{\partial x}$ and $\frac{\partial F}{\partial x}$ when there is no variable $t$, and the letter $x$ is used simultaneously at the bottom two levels. I believe the answer to this question is as follows. 
The fact that $z$ can be expressed in terms of $x$ and $y$ via the formula $z = F(x,y)$ is not relevant in determining the meaning of $\frac{\partial z}{\partial x}$. Just because it is possible to write an expression for $z$ that involves $x$ and $y$ doesn't mean that when you write $\frac{\partial z}{\partial x}$, we must understand $y$ to be held constant. Thus in this case, $x$ should be understood to mean $t$, the lowest-level variable. (Only we must write $\frac{dz}{dx}$ instead of $\frac{\partial z}{\partial x}$, because there is no other variable at the lowest level. If there had been another variable at the lowest level (say $s$), we would have had to write $\frac{\partial z}{\partial x}$.)
On the other hand, when we write $\frac{\partial F}{\partial x}$, the situation is different. The letter $F$ stands for a function of two variables $x$ and $y$, and $x$ is the first of those two variables. In this case we must consider $x$ to mean the middle-level variable.
A: no because $x$ is not dependent of $y$. Moreover $\frac{dF}{dy}$ is not correct because the differential is not total. It should be $\frac{\partial F}{\partial y}=...$
