TITLE: How to explain this quirk of the chain rule?
QUESTION [17 upvotes]: Assume I have a function $f = f(y, \phi(y,x))$ and I want to calculate $\frac{\partial f}{\partial y}$, I use the chain rule to get 
\begin{equation}
\frac{\partial f}{\partial y} = \frac{\partial f}{\partial y} + \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial y} 
\end{equation}
but obviously the $\frac{\partial f}{\partial y}$ represent different things on each side of equality. How do I explain this? I'm guessing it is a notational issue.  

Edit: Just to give some context why this troubles me. Here $x_i$ refers to the ith component of the vector $\mathbf{x}$ in euclidean space. In an acoustic textbook the Lighthill stress tensor $T_{ij}$ is involved in the following identity:
\begin{equation}
\frac{\partial}{\partial x_i} \frac{T_{ij}(\mathbf{y},t-|\mathbf{x}-\mathbf{y}|/c)}{|\mathbf{x}-\mathbf{y}|} = \frac{\frac{\partial T_{ij}}{\partial y_i}}{|\mathbf{x}-\mathbf{y}|} - \frac{\partial}{\partial y_i} \frac{T_{ij}(\mathbf{y},t-|\mathbf{x}-\mathbf{y}|/c)}{|\mathbf{x}-\mathbf{y}|}  \end{equation}
This can only be resolved if the numerator in the term $\frac{\frac{\partial T_{ij}}{\partial y_i}}{|\mathbf{x}-\mathbf{y}|}$ is given a different interpretation...Just try showing this:
Let $t-|\mathbf{x}-\mathbf{y}|/c = \phi(t,\mathbf{x}, \mathbf{y})$
\begin{array}{lcl} 
\frac{\partial}{\partial x_i} \frac{T_{ij}(\mathbf{y},\phi)}{|\mathbf{x}-\mathbf{y}|} & = & \frac{1}{|\mathbf{x}-\mathbf{y}|} \frac{\partial}{\partial x_i}T_{ij}(\mathbf{y},\phi) + T_{ij}(\mathbf{y},\phi) \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{x}-\mathbf{y}|}  \\  
& = &  \frac{1}{|\mathbf{x}-\mathbf{y}|} (\frac{\partial T_{ij}}{\partial \phi}\frac{\partial \phi}{\partial x_i}) + T_{ij}(\mathbf{y},\phi) \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{x}-\mathbf{y}|}\\ 
& = &  -\frac{1}{|\mathbf{x}-\mathbf{y}|} (\frac{\partial T_{ij}}{\partial \phi}\frac{\partial \phi}{\partial y_i}) + T_{ij}(\mathbf{y},\phi) \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{x}-\mathbf{y}|} 
\end{array}
\begin{array}{lcl} 
\frac{\partial}{\partial y_i} \frac{T_{ij}(\mathbf{y},\phi)}{|\mathbf{x}-\mathbf{y}|} & = & \frac{1}{|\mathbf{x}-\mathbf{y}|} \frac{\partial}{\partial y_i}T_{ij}(\mathbf{y},\phi) + T_{ij}(\mathbf{y},\phi) \frac{\partial}{\partial y_i} \frac{1}{|\mathbf{x}-\mathbf{y}|}  \\  
& = &  \frac{1}{|\mathbf{x}-\mathbf{y}|} ( \frac{\partial}{\partial y_i}T_{ij} +\frac{\partial T_{ij}}{\partial \phi}\frac{\partial \phi}{\partial y_i}) - T_{ij}(\mathbf{y},\phi) \frac{\partial}{\partial x_i} \frac{1}{|\mathbf{x}-\mathbf{y}|} 
\end{array}
Adding up the last line from each expression gives the result.

REPLY [3 votes]: In my opinion, Leibniz notation for partial derivatives is terrible: I avoid using it whenever possible, except for a particular usage from differential geometry. (the ambiguity you cite in your question isn't the only problem with it!)
My favorite notation is a variation of $f'$ used for the derivative of a univariate function $f$: the functions $f_1$ and $f_2$ are the functions one would normally write as
$$ f_1(x,y) = \frac{\partial}{\partial x} f(x,y) $$
$$ f_2(x,y) = \frac{\partial}{\partial y} f(x,y) $$
so I would write
$$ \frac{\partial}{\partial y} f(y, \phi(y,x))
= f_1(y, \phi(y,x)) + f_2(y, \phi(y,x)) \phi_1(y,x) $$
Typically, I'm interested in both partials rather than just one partial, and I would use differentials instead of partial derivatives to organize the calculation of all of them at once
$$ \mathrm{d} f(y, \phi(y,x))
= f_1(y, \phi(y,x))\, \mathrm{d}y + f_2(y, \phi(y,x)) \,\mathrm{d}\phi(y,x) = \ldots $$
and when I'm really interested in one partial, I do the same thing, except work in the setting where I've set $\mathrm{d}x=0$. (assuming the partial I really mean to use is the one where $x$ is held constant)

In the differential geometry setting, in my opinion there is no ambiguity:
$$ \frac{\partial}{\partial x^i} f(x^i, g(y^i, x^i)) $$
has only one reasonable meaning: applying the tangent vector $\partial/\partial x^i$ to the scalar field $f(x^i, g(y^i, x^i))$ in the $i$-th coordinate direction. In my opinion, you wouldn't use that notation when you wanted the derivative of $f$ with respect to its first argument.
(although this example has two sets of independent variables: which makes me again dislike using Leibniz notation for it)