= Solution
The <regular level set theorem> says that for a $C^1$ map $g:U\subseteq\mathbb R^n\to\mathbb R^r$, if $Dg$ has rank $r$ on $g^{-1}(c)$, then $g^{-1}(c)$ is an $(n-r)$-dimensional <submanifold> and
$$
T_p(g^{-1}(c))=\ker Dg(p).
$$
Here $r=1$ and the given rank condition makes the zero set a two-dimensional <submanifold>. It is nonempty because $g(0)=f(0)=0$. Part (c) gives
$$
Dg(p)(1,1,1)=g_x(p)+g_y(p)+g_z(p)=0.
$$
Consequently
$$
\boxed{(1,1,1)\in\ker Dg(p)=T_pM\quad\text{for every }p\in M.}
$$
There is also a direct curve proof: $g(p+t(1,1,1))=g(p)=0$, so the entire line through $p$ in this direction lies in $M$. Its <velocity> at zero is exactly the required <tangent vector>.
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