I was reading the section of calculus of variations in Taylor's Classical mechanics and he went over some examples. The first being:
When he reaches the portion $\frac{d}{dx}\frac{\partial f}{\partial y'}=0$, it makes that he claims $\frac{\partial f}{\partial y'}$ is a constant. In the next example (the brachistocrhone),
in taking $\frac{d}{dy}\frac{\partial f}{ \partial x'}$ he gets $\frac{x'}{\sqrt{y}(1+x'^2)^{1/2}}$ where he completely ignores the y in the equation. How can he do that?
Since $f$ does not depend at all on $x$, he has $$ 0 = \frac{\partial f}{\partial x} = \frac{d}{dy}\frac{\partial f}{\partial x'}.$$
Something with zero derivative is constant, so we can just ignore the $d/dy$ and write $$\frac{\partial f}{\partial x'} = C.$$ Working out the $x'$ derivative, $$ \frac{\partial f}{\partial x'} = \frac{ x'\sqrt{y}}{y\sqrt{x'^2+1}}. $$
Squaring and canceling gives his equation.