Partial derivative of a function with respect to its derivative While going through some introductory notes on Lagrangian and Hamiltonian Mechanics, I was a bit surprised by the notion of a functional
$$         J(y):=\int_{a}^{b} F\left(x, y, y_{x}\right) \mathrm{d} x $$
and how it depends on $x, y$ and $y_x$. Given that $y = y(x)$, it follows that $ F = F(x)$, and I didn't understand the need to show an explicit dependence on $ y $ and $y_x$. The confusion that I specifically have is with respect to a solution of the brachsitochrome problem, 
$$
\begin{aligned}
\frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}-y_{x} \frac{\partial}{\partial y_{x}}\left(\frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}\right) &=c \\
\frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}-\frac{y_{x}}{\left(y_{1}-y\right)^{\frac{1}{2}}} \cdot \frac{\partial}{\partial y_{x}}\left(\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}\right) &=c
\end{aligned} $$
What does it mean to take the partial of a function with respect to its derivative, that is, what does $ 
\frac{\partial y}{\partial y_{x}} $ mean and why is it equal to zero? In other words, why were able to bring $ (y_1 - y)^{\frac{1}{2}} $ outside the $\frac{\partial}{\partial y_x}$ ?
 A: This notation can be defined in the following more rigorous way
$$
\partial_{y_x}F(x,y,y_x) := (\partial_3F)(x,y,y_x)
$$
where $\partial_3$ is the derivation with respect to the third variable of $F$. If you prefer, you can also write $F: (x,Y,Z)↦F(x,Y,Z )$ and then
$$
\partial_{y_x}F(x,y,y_x) = (\partial_ZF)(x,y,y_x).
$$
In pratice, you first differentiate assuming your functions inside are independent variables, and then you evaluate.
A: An ODE as
$$
y''y'+3y'^2+x y=0
$$
can be represented as 
$$
F(x,y,y',y'')=0
$$
or
$$
F(x,u_1,u_2,u_3)=0,\ \ \cases{u_1=y\\
u_2=\frac{d u_1}{dx}\\
u_3=\frac{d u_2}{dx}}
$$
now if we want to derive it regarding $x$ we can proceed as follows
$$
\frac{d F}{dx}=\frac{\partial F}{\partial x}\frac{dx}{dx}+\frac{\partial F}{\partial u_1}\frac{d u_1}{dx}+\frac{\partial F}{\partial u_2}\frac{d u_2}{dx}+\frac{\partial F}{\partial u_3}\frac{du_3}{dx}
$$
or
$$
\frac{d F}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}+\frac{\partial F}{\partial y'}\frac{dy'}{dx}+\frac{\partial F}{\partial y''}\frac{dy''}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y''+\frac{\partial F}{\partial y''}y'''
$$
now applying to our example
$$
\cases{
\frac{\partial F}{\partial x}=y\\
\frac{\partial F}{\partial y}=x\\
\frac{\partial F}{\partial y'}=y''+6y'\\
\frac{\partial F}{\partial y''}=y'
}
$$
and the result
$$
\frac{dF}{dx}=y + x y'+(y''+6y')y''+y'y'''=0
$$
