I'm trying to prove the relationship between total derivative and partial derivatives. Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
Theorem 1: Suppose $F$ is a normed vector space and $f: X \subseteq \mathbb{R}^{n}\rightarrow F$ is differentiable at $a$. Then $$ \partial f\left(a\right) (h)=\sum_{j=1}^{n} \partial_{k} f\left(a\right) \cdot h_k $$ for all $h=\left(h_1, \ldots, h_n\right) \in \mathbb{R}^{n}$.
and
Theorem 2: Suppose $X$ is a metric space and $f=\left(f_{1}, \ldots, f_{n}\right): X \rightarrow \mathbb{K}^{n}$. Then $f$ is differentiable at $a$ if and only if $f_j$ is differentiable at $a$ for all $j = \overline{1,n}$. Moreover, $$\partial f\left(a\right)=\left(\partial f_{1}\left(a\right), \ldots, \partial f_{n} \left(a\right)\right)$$
My attempt:
Theorem 1: Let $\{e_k \mid 1 \le k \le n\}$ be the standard basis of $\mathbb R^n$. We have $$\begin{aligned} \partial f\left(a\right) (h) = \partial f\left(a\right) \left(\sum_{j=1}^n h_k e_k \right) = \sum_{j=1}^n \partial f(a) (e_k) \cdot h_k = \sum_{j=1}^n \partial_k f(a) \cdot h_k \end{aligned}$$
Theorem 2: Let $(y_1, \ldots,y_m) = \partial f(a) (h) \in \mathbb K^n$. We have $$\begin{aligned} \lim_{h \to 0} \frac{f(a+h)-f(a)- \partial f(a) (h)}{\|h\|} &= 0 \\ \iff \lim_{h \to 0} \frac{(f_1(a+h),\ldots,f_n (a+h))-(f_1(a),\ldots,f_n (a)) - (y_1, \ldots,y_n)}{\|h\|} &= 0 \\ \iff \lim_{h \to 0} \frac{(f_1(a+h)-f_1(a)-y_1, \ldots, f_n(a+h)-f_n (a)-y_n)}{\|h\|} &=0 \\ \iff \lim_{h \to 0} \left ( \frac{f_1(a+h)-f_1(a)-y_1}{\|h\|},\ldots, \frac{f_n (a+h)-f_n (a)-y_n}{\|h\|} \right) &=0 \\ \iff \left ( \lim_{h \to 0} \frac{f_1(a+h)-f_1(a)-y_1}{\|h\|},\ldots, \lim_{h \to 0} \frac{f_n (a+h)-f_n (a)-y_n}{\|h\|} \right) &=0 \\ \iff \forall j = \overline{1,n}: \lim_{h \to 0} \frac{f_1(a+h)-f_1(a)-y_1}{\|h\|} &=0 \\ \iff \forall j = \overline{1,n}: \partial f_j(a) (h) &= y_j \\ \iff (\partial f_1(a),\ldots, \partial f_n(a)) &= \partial f(a) \end{aligned}$$