= Solution
Assume the <holonomic constraints> have <linearly independent> <gradients>, so their common level <set> is locally a smooth constraint <manifold> parametrized by $q_1,\ldots,q_n$. The multiplier form of the <Euler-Lagrange equations> is
$$
\frac d{dt}L_{\dot x_A}-L_{x_A}
=\sum_\alpha\lambda_\alpha\partial_A f_\alpha.
$$
At fixed time $\sum_A\partial_A f_\alpha\,\partial_i x_A=0$. The <chain rule> for the reduced <Lagrangian> gives
$$
\frac d{dt}\mathcal L_{\dot q_i}-\mathcal L_{q_i}
=\sum_A\left(\frac d{dt}L_{\dot x_A}-L_{x_A}\right)\partial_i x_A=0.
$$
Conversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. <Linear independence> of the constraint <gradients> implies that it is a linear combination of those <gradients>, giving the required <Lagrange multipliers>. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.
For the bead, $\dot y=f'(x)\dot x$, so
$$
\boxed{\mathcal L=\frac12(1+f'(x)^2)\dot x^2},\qquad
(1+f'^2)\ddot x+f'f''\dot x^2=0.
$$
The autonomous <conservation of energy> gives $E=\tfrac12(1+f'^2)\dot x^2$. If $E>0$, the motion has a fixed sign $s=\operatorname{sgn}\dot x$ and
$$
\boxed{t=t_*+\frac{s}{\sqrt{2E}}\int_{x_*}^{x}\sqrt{1+f'(u)^2}\,du}.
$$
The <integral> is the signed <arc length> divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a <function> of its unchanging position is not meaningful.
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