Here are two theorems from T. Apostol's book on analysis.
Since the left-hand sides of both displays are the same (up to replacing $u$ with $v$ or vice versa), I believe their right-hand sides are equal too (provided in $(6)$ one replaces $u$ with $v$). So we have the following equality: $$f'(c;v)=\sum_{k=1}^nv_k D_k f(c),$$ where $v=(v_1,\dots,v_n)^t\in \mathbb R^n$.
Is there a more direct way of proving this? I tried to expand the quotient in the definition of the directional derivative $f'(c;v)$ but didn't arrive at anything reasonable.
Also, am I correct in saying that both these theorems are generalized versions of Theorem 9.17 in Baby Rudin? (See Theorem 9.17 from Baby Rudin)
I would show the equality by using the abstract characterization of the derivative. That is, for some function $f\colon \Omega \to \mathbb{R}^n$ mapping an open domain $\Omega \subseteq R^m$ into the $n$-dimensional euclidean space, we call a linear function $T\colon \mathbb{R}^m \to \mathbb{R}^n$ the derivative of $f$ in $c\in \Omega$ iff $f(c+h) - f(c) - T(h) = o(|h|)$. The later means that $$ \lim_{h \to 0} \frac{|f(c+h) - f(c) - T(h)|}{|h|} = 0.$$ Indeed, it can be easily shown that such a linear function is unique if it exists.
We now want to show that $$f'(c;v)=^?\sum_{k=1}^nv_k\, D_k f(c).$$ However, the right hand side equals the (total) derivative of $f$ at $c$, and the left hand side fulfills $$\lim_{t \to 0^+} \frac{|f(c+t v) - f(c) - f'(c;v) t|}{|t|} = 0$$ by the very definition of the directional derivative. Since the directional derivative is unique as well, it follows that the putative equality indeed holds.