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Suppose $f=f(x,y(x))$.

Then applying the chain rule we get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}$.

From this it seems that it always holds that $\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=0$.

Where's the mistake?

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    It doesn't matter here, since you only have one variable. Your error is in applying the chain rule to the outermost function of two variables.2012-08-08

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As usual when there's confusion about partial derivatives, everything is readily cleared up if we remedy the deficiency in our notation for them by marking which variables are being held fixed:

$ \def\part#1#2#3{\left.\frac{\partial #1}{\partial #2}\right|_{#3}} \part fxz=\part fxy\part xxz+\part fyx\part yxz=\part fxz=\part fxy+\part fyx\part yxz\;, $

so there's no such implication, since

$ \part fxz\ne\part fxy\;, $

unless of course you choose $z=y$, in which case indeed

$ \part yxz=\part yxy=0\;. $

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    Finally, after 3 months, someone have an answer to what is basically the [same question](http://math.stackexchange.com/questions/1161528/nontrivial-chain-rule-diagrams-how-to-write-chain-rule-for-them-and-is-there-im) I have asked 3 months ago. I should favourite it and check my calculations now2015-05-25