For the following theorem:
Spivak stated,
Although the chain rule was used ... it could easily have been eliminated.
I believe there is a proof without invoking the auxiliary function $h$ - but I don't see how we can show $D_jf(a)$ exists, moreover, even if it exists, I couldn't show the derivative of $f$ is of the desired form without using MVT...
Denote by $A$ the differential of $f$ at $x = a$. Denote the projection onto the $i$-th coordinate of $\mathbb{R}^m$ by $\pi_i \colon \mathbb{R}^m \rightarrow \mathbb{R}$. This is a continuous and linear map and so
$$ D_j f^i(a) := \lim_{t \to 0} \frac{f^i(a + te_j) - f^i(a)}{t} = \lim_{t \to 0} \frac{\pi_i (f(a + te_j)) - \pi_i f(a))}{t} = \lim_{t \to 0} \pi_i \left( \frac{f(a + te_j) - f(a)}{t} \right) = \pi_i \left( \lim_{t \to 0} \frac{f(a + te_j) - f(a)}{t} \right).$$
By definition of the differential,
$$ \lim_{v \to 0} \frac{f(a + v) - f(a) - Av}{\| v \|} = 0. $$
Choosing $v = te_j$ we have
$$ \lim_{t \to 0} \frac{f(a + te_j) - f(a) - A(te_j)}{t} = \lim_{t \to 0} \frac{f(a + te_j) - f(a)}{t} - Ae_j = 0 \implies \\ \lim_{t \to 0} \frac{f(a + te_j) - f(a)}{t} = Ae_j.$$
and so $D_j f^i(a) = \pi_i(Ae_j) = e_i^T A e_j$ is the $(i,j)$-entry of the matrix $A$.