= Solution
Let $g_1=x^2+y^2+z^2-4$ and $g_2=(x-3)^2+y^2+z^2-4$. Subtracting the two constraint equations gives $-6x+9=0$, hence $x=3/2$ and $y^2+z^2=7/4$. The feasible <set> is therefore a <circle>.
For <Lagrange multipliers> $\lambda,\mu$, the stationarity equation $\nabla f=\lambda\nabla g_1+\mu\nabla g_2$ becomes
$$
0=2\lambda x+2\mu(x-3),\qquad
1=2(\lambda+\mu)y,\qquad
0=2(\lambda+\mu)z.
$$
At $x=3/2$, the first equation gives $\lambda=\mu$. The second ensures $\lambda+\mu\ne0$ and $y\ne0$, so the third gives $z=0$. The constraints then force $y=\pm\sqrt7/2$. Both points do satisfy the multiplier equations, taking $\lambda=\mu=1/(4y)$.
There are no missing singular constraint points: on the feasible <circle>, the two constraint <gradients> $(3,2y,2z)$ and $(-3,2y,2z)$ are <linearly independent>, since $y^2+z^2=7/4>0$. Thus the complete list of constrained <critical points> and <critical values> is
$$
\boxed{(x,y,z)=\left(\frac32,\pm\frac{\sqrt7}{2},0\right),\qquad f=\pm\frac{\sqrt7}{2}.}
$$
The bound $|y|\leq\sqrt7/2$ on the feasible <circle> shows that the positive value is the global <maximum> and the negative value is the global <minimum>.
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