Consider a guide for marbles whose profile locally coincides with a function $f(x)$, for example $f (x) = - \frac{1}{2} x ^ 4 + x ^ 3 + x ^ 2-2x + 1.$
Suppose that the reference system is chosen in such a way that $f(x)$ coincides with the altitude; then we know that $mgf(x)$ represents the gravitational potential energy $U(x)$ of a marble placed at $x$. Is it possible to write the equation of motion of a marble, fixed the initial $x,y$ coordinates?
the position x,y of the marble is:
$$x\mapsto x\\ y\mapsto f(x)$$
from here you can obtain the kinetic energy T
$$T=\frac m2(\dot x^2+\dot y^2)=\frac m2\left[\dot x^2+ \left(\frac{\partial f(x)}{\partial x}\dot x\right)^2\right]+\frac{I_M}{2}\,\dot\varphi^2$$
and the potential energy
$$U=-m\,g\,f(x)$$
apply EL with the roll condition $~\dot\varphi=\frac{\dot x}{r}~$ to obtain the equation of motion
$${\ddot x}+{\frac {{r}^{2} \left( {\frac {d}{dx}}f \left( x \right) \right) m \left( {{\dot x}}^{2}{\frac {d^{2}}{d{x}^{2}}}f \left( x \right) +g \right) }{m{r}^{2}+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}m{r}^{2}+I_{{M}}}} =0$$
Edit
the roll condition should be $~{\dot\varphi}=\frac{v_t}{r}~$
where $~v_t~$ is the path tangential velocity.
$$v_t=\pm{\dot x}\,\sqrt {1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}} $$
thus the EOM $$\ddot x+ \left[ \begin {array}{c} {\frac { \left( {\frac {d}{dx}}f \left( x \right) \right) \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) {{\dot x}}^{2}}{1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}}}+{\frac { \left( {\frac {d}{dx}}f \left( x \right) \right) m\,g\,{r}^{2}}{ \left( 1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2} \right) \left( m{r}^{2}+{\it I_M} \right) }} \end {array} \right] =0$$
$$y = f(x)$$ means that one can consider parametrization \begin{align} &x = q\\ &y = f(q) \end{align} Let us assume the marble rolls along the curve without friction and it is always in contact with the curve $y = f(x)$, whose radius of curvature is bigger the the marble's radius. A unit vector, tangent to the curve at a point $(q, \, f(q))$ is $$\left(\frac{1}{\sqrt{1+f'(q)^2\,}\,}\,,\,\,\, \frac{f'(q)}{\sqrt{1+f'(q)^2\,}\,}\right)$$ and hence, a unit vector normal (i.e. perpendicular) to the curve at point $(q, \, f(q))$ is $$\left(-\,\frac{f'(q)}{\sqrt{1+f'(q)^2\,}\,}\,,\,\,\, \frac{1}{\sqrt{1+f'(q)^2\,}\,}\right)$$ The position of the marble's center of mass is \begin{align} &x \,=\, q\,-\,\frac{r\,f'(q)}{\sqrt{1+f'(q)^2\,}\,}\\ &y \,=\,f(q)\,+\,\frac{r}{\sqrt{1+f'(q)^2\,}\,} \end{align} and its velocity can be expressed as
\begin{align} &\frac{dx}{dt}\,=\, \left(\, 1 \,-\,\frac{r\,f''(q)}{\sqrt{1+f'(q)^2\,}\,} \,+\, \frac{r\,f'(q)^2f''(q)}{\big(\,1 + f'(q)^2\,\big)^{\frac{3}{2}}}\,\right) \frac{dq}{dt}\\ &\frac{dy}{dt} \,=\, \left(\,f'(q)\,-\,\frac{r\,f'(q)f''(q)}{\big(\,1 + f'(q)^2\,\big)^{\frac{3}{2}}} \,\right) \frac{dq}{dt} \end{align}
Abbreviate
\begin{align} &V(q) \,=\, 1 \,-\,\frac{r\,f''(q)}{\sqrt{1+f'(q)^2\,}\,} \,+\, \frac{r\,f'(q)^2f''(q)}{\big(\,1 + f'(q)^2\,\big)^{\frac{3}{2}}}\\ &W(q) \,=\,f'(q)\,-\,\frac{r\,f'(q)f''(q)}{\big(\,1 + f'(q)^2\,\big)^{\frac{3}{2}}} \end{align}
and thus
\begin{align} &\frac{dx}{dt}\,=\, V(q) \, \frac{dq}{dt}\\ &\frac{dy}{dt} \,=\, W(q) \, \frac{dq}{dt} \end{align}
The angular velocity of the marble from rotation is related to the arclength traversed by the marble as follows: $$r\,\frac{d\varphi}{dt}\,=\, \frac{ds}{dt} \,=\, \sqrt{1 + f'(q)^2\,}\,\frac{dq}{dt}$$
which can be denoted by $$S(q) \,=\, \sqrt{1 + f'(q)^2\,}$$ $$r\,\frac{d\varphi}{dt} \,=\, S(q)\,\frac{dq}{dt}$$
The kinetic and potential energy of the marble are
\begin{align} T \,=\, \frac{m}{2} \left(\,\Big(\frac{dx}{dt}\Big)^2 +\, \Big(\frac{dy}{dt}\Big)^2\,\right) \,+\, \frac{I}{2} \Big(\frac{d\varphi}{dt}\Big)^2 \end{align} $$U \,=\, gm\,y$$
Express the kinetic and potential energy in terms of the coordinate $q$: $$T(q) \,=\, m\Big(\,V(q)^2 +\, W(q)^2\,\Big) \,+\, \frac{I}{r^2} \,S(q)^2$$ \begin{align} T\,=\,\frac{1}{2}\,T(q) \,\Big(\frac{dq}{dt}\Big)^2 \,=\, \left(\,\frac{m}{2} \Big(\,V(q)^2 +\, W(q)^2\,\Big) \,+\, \frac{I}{2r^2} \,S(q)^2 \right) \Big(\frac{dq}{dt}\Big)^2 \end{align} $$U(q) \,=\, gm\,\left(\,f(q)\,+\,\frac{r}{\sqrt{1+f'(q)^2\,}\,}\,\right)$$
Now, to form the eqation of motion, one can go about it in, say, two ways. One can write the Lagrangian of the system $$L \,=\,\frac{1}{2}\, T(q) \,\Big(\frac{dq}{dt}\Big)^2 \,-\, U(q)$$ and derive the Euler-Legrange second order differential equation: $$\frac{d}{dt} \left(\,T(q) \,\Big(\frac{dq}{dt}\Big)\,\right) \,=\, \frac{1}{2}\, T'(q) \,\Big(\frac{dq}{dt}\Big)^2\,-\, U'(q)$$ which can be rewritten in the form $$T(q) \,\frac{d^2q}{dt^2} \, + \, \frac{1}{2}\, T'(q) \,\Big(\frac{dq}{dt}\Big)^2 \,+\, U'(q) \,=\, 0$$ You could also use the conservation of energy, i.e. $$\frac{1}{2}\, T(q) \,\Big(\frac{dq}{dt}\Big)^2 \,+\, U(q) \,=\, E_0$$ where $E_0 \,=\, U(q_0)$ is the constant determined by the initial position of the marble with zero initial velocity: $x_0 = q_0$ and $y_0 = f(q_0)$. After solvling for the derivative of $q=q(t)$ $$\frac{dq}{dt}\,=\, \pm\, \sqrt{\,\frac{2\,E_0 \,-\, 2\,U(q)}{T(q)} \,}$$ The expressions for the functions involved in the right hand side of the equation are fairly heavy. One can achieve a simplification by setting $r \to 0$.
