= Solution
Use
$$
L(x,y,z,\lambda)
=x^2+y^4+z^6-\lambda(x+2y+3z-6).
$$
Stationarity gives
$$
2x=\lambda,
\qquad
4y^3=2\lambda,
\qquad
6z^5=3\lambda.
$$
Putting $t=\lambda/2$, this becomes
$$
x=t,\qquad y=t^{1/3},\qquad z=t^{1/5},
$$
and the constraint requires
$$
t+2t^{1/3}+3t^{1/5}=6.
$$
The left side is strictly increasing, and $t=1$ solves the equation. Hence
$$
\boxed{(x,y,z)=(1,1,1)},
\qquad
\boxed{f_{\min}=3},
\qquad
\lambda=2.
$$
The objective is convex and the constraint is affine. Its tangent-plane inequality at $(1,1,1)$ gives, for every feasible $(x,y,z)$,
$$
f(x,y,z)\geq3+(2,4,6)\cdot(x-1,y-1,z-1)=3,
$$
so the Lagrange point is globally optimal. Moreover, at the dual value $\lambda=2$, the infimum of the <Lagrangian function in constrained optimization> is attained at the same point and equals three. The primal and dual values coincide, so strong duality holds.
For the value <function>, the multiplier convention above gives the <derivative of a constrained value function>
$$
\phi'(b)=\lambda(b).
$$
At $b=6$, therefore,
$$
\boxed{\phi'(6)=2}.
$$
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