When reading a book related to the calculus of variations, I came across a proof about partial differentiation that confused me a lot. On a given surface $G(x,y,z) = 0$, we have to find a space curve lying on that surface where $x,y,z$ are all functions of a parameter $t$, that gives a stationary value to the integral of the form $$\int_{t_1}^{t_2} f(\dot{x},\dot{y},\dot{z}) dt.$$ The proof begins with writing $z = g(x,y)$, and thus \begin{equation}\dot{z} = \frac{\partial g}{\partial x}\dot{x} + \frac{\partial g}{\partial y}\dot{y}. \end{equation} The problem is that I failed to understand the following two statements:
The Euler-Lagrange equations becomes $$\frac{d}{dt}(\frac{\partial f}{\partial \dot{x}}+\frac{\partial f}{\partial \dot{z}}\frac{\partial g}{\partial x}) - \frac{\partial f}{\partial \dot{z}}\frac{\partial \dot{z}}{\partial x} = 0$$ and a similar one for $y$, which I am unable to derive from its simpler version $$\frac{d}{dt}(\frac{\partial f}{\partial \dot{x}}) - \frac{\partial f}{\partial x} = 0$$ when there are no $z$ to begin with.
The proof said that follow from the equation for $\dot{z}$ above it can be seen that $$\frac{\partial{\dot{z}}}{\partial x} = \frac{d}{dt}(\frac{\partial g}{\partial x})$$ and similar for $y$, which causes me to be confused as $x$ itself is a function of $t$, and mixing partial derivative and total derivative here make me unable to comprehend how can it be established. I wonder whether I can use Clairaut's Theorem but it requires all derivatives to be partial. Any help would be appreciated.
This is just the chain rule. We have a constraint $$z~=~g(x),\tag{1}$$ where $x=(x^1,x^2)$. The $t$-derivative of the constraint (1) becomes $$\dot{z}~\stackrel{(1)}{=}~\sum_i \dot{x}^i \frac{\partial g}{\partial x^i} . \tag{2}$$ The Lagrangian therefore reads $$L(x,\dot{x})~:=~f\left(\dot{x},\sum_i \dot{x}^i \frac{\partial g}{\partial x^i}\right).\tag{3}$$ Euler-Lagrange (EL) equations become $$ \frac{d}{dt} \left(\frac{\partial f}{\partial \dot{x}^k}+\frac{\partial f}{\partial \dot{z}} \frac{\partial g}{\partial x^i}\right) ~\stackrel{(3)}{=}~ \frac{d}{dt} \frac{\partial L}{\partial \dot{x}^k} ~=~ \frac{\partial L}{\partial x^k} ~\stackrel{(3)}{=}~\frac{\partial f}{\partial \dot{z}}\sum_i \dot{x}^i \frac{\partial^2 g}{\partial x^i\partial x^k} .\tag{4}$$ In eq. (4) the notation $\frac{\partial f}{\partial \dot{z}}$ just means partial differentiation of $f$ wrt. its last entry.