I recall (a version of) the multivariate chain rule: Let $z = f(x,y)$ be a differentiable function of $x$ and $y$, where $x = g(t)$ and $y = h(t)$ are themselves differentiable functions of $t$. Then \begin{eqnarray} \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}. \quad \quad \quad (1) \end{eqnarray} My confusion has arisen from asking, what if $y(t) = t$? Blindly following (1), we seem to have, \begin{eqnarray} \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial t} \quad \quad \quad (2) \end{eqnarray} which is wrong, of course.. ((2) $\Leftrightarrow \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}=0$, but a counterexample: take $f(x,t)=x$ with $x(t)=t$, so $\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}=1\neq0$).
So, to confirm, is (2) is incorrect? Why doesn't (1) work with just putting $y(t)=t$? And, most importantly, what should the correct version of (2) be?
Many thanks, and, sorry for an elementary question... I've just forgotten my multivariate calculus!