Chain rule for multivariable functions confusion 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?

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