Conservation of total energy for a system with holonomic constraints Consider a system with generalized coordinates $u_1, u_2$ and $u_3$ such that $u_1$ and $u_2$ are dependent through the following holonomic constraint
\begin{equation}
G(u_1, u_2)=0.
\end{equation}
It is also given that generalised force corresponding to each coordinate is zero. 
Kinetic energy of the system is given by 
\begin{equation}
T(u_1, u_2, u_3, \dot{u}_1,\dot{u}_2, \dot{u}_3)=\frac{1}{2}\dot{\bf{u}}^TD(\textbf{u})\dot{\textbf{u}}
\end{equation} 
where $\textbf{u}=[u_1, u_2, u_3]^T$ and $D(\textbf{u})$ is positive definite for all $\textbf{u}$.
The potential energy of the system is given by a function $U(\textbf{u})$. 
Will the total energy $T+U$ be constant?
 A: *

*Assume 


*

*(i) that the kinetic term $T$ is quadratic in generalized velocities $\dot{\bf u}$; 

*(ii) that the potential term $U$ is independent of the generalized velocities $\dot{\bf u}$; and 

*(iii) that the Lagrangian $L=T-U$ does not depend explicitly on time. 


*Case without holonomic constraints. The energy $h=\dot{\bf u}\cdot \frac{\partial L}{\partial \dot{\bf u}}-L=T+U$ is conserved because the Lagrangian $L=T-U$ does not depend explicitly on time, cf. e.g this Phys.SE post. 

*Case with holonomic constraints $G({\bf u})\approx 0$ without explicit time dependence. 


*

*(i) Either we can formally eliminate variables such that there are no holonomic constraints left; or 

*(ii) alternatively, we can introduce Lagrange multipliers, which we add to the list of variables ${\bf u}$, and add terms of the form 'Lagrange multiplier times $G$' to the potential term $U$. (The notion of potential energy will be unaltered on-shell.)
In both cases, the form 1 is maintained, and we can apply the conclusion from section 2: Energy is still conserved.
A: your example
you have three 3 degree of freedom $u_1,u_2,u_3$ (not generalized coordinate) and 1 constraint equation $g(u_1,u_2)=0$ so you have 2 generalized coordinate .
I see two cases:
I) form the constraint equation $g(u_1,u_2)=0$ you can obtain explicit for example $u_2=u_2(u_1)$   so your position vector (mechanical system) is:
$$\vec{r}=\vec{r}(u_1,u_3)$$
$$\vec{v}=\vec{\dot{r}}=\frac{\partial \vec{r}}{\partial u_1}\dot{u}_1+\frac{\partial \vec{r}}{\partial u_3}\dot{u}_3$$
$\Rightarrow$
$$T=T(u_1,\dot{u}_1,u_2\,\dot{u}_2)=m\,\frac{1}{2}\,\vec{v}^T\,\vec{v}$$
$$U=U(u_1,u_3)$$
so:
$$\frac{d}{dt}(T+U)=0$$
The kinetic energy plus potential energy is conserved
II) If you can't eliminate one of the degree of freedom  from the constraint equation then:
from:
$$g(u_1,u_2)=0\quad \Rightarrow\quad \frac{\partial g}{\partial u_1}\dot{u}_1+\frac{\partial g}{\partial u}_2\dot{u}_2=0$$
so $$\dot{u}_2=\dot{u}_2(u_1,u_2,\dot{u}_1)\quad u_2=\int \dot{u}_2\,dt$$
so again like case I the  kinetic energy plus potential energy is conserved
