Where do you consider the motion along the bowl? It will lead to accelerations both in vx and vy, to follow the shape.I'll try to start from scratch - with energy conservation, but it is possible to avoid this word if necessary.
If the object is at rest at position ##x_0>0##, it has a total energy of ##gx_0^2##, setting its mass and a meter to 1.
At position x, ##0<x<x_0##, energy conservation leads to ##v^2=2g(x_0^2-x^2)##.
The derivative of the parabola there is 2x, therefore ##v_x=\frac{1}{2x}v_y##. It follows that ##v^2 = v_x^2 + v_y^2 = v_x^2 (1+4x^2)##. Combining both,
$$\frac{dx}{dt}= v_x = \sqrt{\frac{v^2}{1+4x^2}} = \sqrt{\frac{2g(x_0^2-x^2)}{1+4x^2}}$$
The sign is arbitrary and depends on the current part of the oscillation.
This leads to ugly elliptic integrals. It can be written as
$$\frac{dt}{dx}=\sqrt{\frac{1+4x^2}{2g(x_0^2-x^2)}}$$
While this gives elliptic integrals as well, it allows to determine t(x) via numerical integration.
g=10 and x
0=1
leads to a period of T=2.36s.
For large x
0 (>>1), most of the parabola is like a free fall, and the period should approach 4 times this free-fall time. As an example, x=sqrt(500) leads to T=40.06, whereas a free fall would give T=40s.
For small x
0 (<<1), the deflection of the bowl is negligible, and we get a harmonic oscillator. The numerator for dt/dx can be approximated as 1, and the period approaches ##T \approx \frac{2\pi}{\sqrt{2g}}##.
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