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?