How to derive Euler-Lagrange equation for isochronous curve of Leibniz in terms of $t$, $x(t)$, $\dot{x}(t)$? 
According to this source, "An isochronous curve of Leibniz is a curve such that if a particle comes down along it by the pull of gravity, the vertical component of the speed is constant, when the gravitational field is supposed to be uniform."

Suppose the curve is given by $(x(t),y(t))$. I am attempting to for solve the function $x(t)$ using Lagrangian methods as follows.
The vertical component of speed is constant, $\dot{y}(t)=v_y$. Assuming mass is $m=1$, our kinetic energy is $K=\frac{1}{2}(\dot{x}^2+v_y^2)$ and our potential energy is $U=gv_y t$.
Then we can formulate a Lagrangian,
$$L(t,x(t),\dot{x}(t))=\frac{1}{2}(\dot{x}^2+v_y^2)-gv_y t$$
and we can calculate that $\frac{\partial L}{\partial x}=0$ and $\frac{\partial L}{\partial \dot{x}} = \dot{x}$
so the Euler-Lagrange equation implies
$$\frac{d}{dt}\dot{x}(t)=0.$$
But this is obviously incorrect. The correct answer is actually
$$x(t) = \frac{2}{3}\sqrt{gv_yt^3}.$$ I also know (from using the Newtonian method) that the solution arises from the fact that $\dot{x}=\frac{gv_y}{\ddot{x}}$, and I suspect the Lagrangian should result in something similar.
Where have I erred in this methodology? Is it salvageable?
 A: according to your source :
$$\dot{y}^2=2\,g\,x$$
thus:
$$T=\frac 12 m\,(\dot{x}^2+2\,g\,x)$$
$$U=g\,x$$
with E.L: you obtain
$$\ddot{x}=0$$
$\Rightarrow$
$$x(t)=v_0\,t$$
$$\dot{y}=\sqrt{2\,g\,v_0\,t}~,y(t)=\frac 23\,{t}^{\frac 32}\sqrt {2}\sqrt {g}\sqrt {{v_0}}$$
so you get the same result that given in your source
$$y^2={\frac {8}{9}}\,{\frac {g{x}^{3}}{{{v_0}}^{2}}}$$
A: *

*OP's question has a direct analog in the brachistochrone problem. There one could also ask for a Lagrangian for the point mechanical problem, cf. e.g. this Math.SE post. However, this would be overkill as the point mechanical problem is readily solved by energy conservation alone:
$$ v_x^2+v_y^2-v_0^2~=~2gy,\qquad v_y~=~v_0.\tag{1}$$
(Here the $y$-axis points downwards & we assume for simplicity vertical initial velocity.)


*Instead a candidate for a variational problem concerns the choice of background geometry. However even this problem is too simple to bother with variational settings. The  isochronous curve of Leibniz is the solution to the following ODE:
$$ v_0~\stackrel{(1)}{=}~v_y~=~\frac{dy}{dx}v_x~\stackrel{(1)}{=}~\frac{dy}{dx}\sqrt{2gy}. \tag{2}$$
