= Solution
{c}
At a regular point of the <optimization constraint>, meaning $\nabla g\ne0$, the necessary <Lagrange multiplier> condition is
$$
\boxed{\nabla f=\lambda\nabla g\quad\text{for some }\lambda\in\mathbb R.}
$$
Indeed, every allowed <tangent vector> $t$ to the <level set> $g=0$ satisfies $\nabla g\cdot t=0$. At a constrained <extremum>, the <directional derivative> $\nabla f\cdot t$ must vanish too. Thus both <gradients> are normal to the same <tangent line> and are parallel. This includes $\nabla f=0$. At a singular point with $\nabla g=0$, the displayed condition need not be necessary; those points require a separate examination.
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