Show that the proof of Theorem $6.2$ goes through if we assume merely that the partials $D_jf$ exist in a neighborhood of $a$ and are continuous at $a$ then $f$ is differentiable in $a$.
Theorem $6.2.$ Let $A$ be open in $\mathbb{R}^m$. Suppose that the partial derivatives $D_jf_i(x)$ of the component functions of $f$ exist at each point $x$ of $A$ and are continuous on $A$. Then $f$ is differentiable at each point of $A$.
I am reading the book "Analysis on manifolds" by Munkres and in the proposed exercises this problem is found, I have already thought a lot about this problem and I know that there are many related posts on this site about this, but I would like someone to explain to me how this can be demonstrated and how this can be applied to a particular problem by means of an example, thank you very much.
The idea here is straightforward, but notationally messy.
First notice that it is sufficient to deal with $f:A \to \mathbb{R}$.
It is easy to guess the candidate derivative as $h \mapsto \sum_k {\partial f(x) \over \partial x_k} h_k$ (the sum is taken over $k=1,...,n$).
Now you need to find an appropriate estimate of $\|f(x+h)-f(x) - \sum_k {\partial f(x) \over \partial x_k} h_k\|$.
Let $\phi_k(h) = (h_1,\cdots,h_k,0,\cdots)$, with $\phi_0(h) = 0$.
We can write $f(x+h)-f(x) = \sum_k f(x+\phi_k(h))-f(x+\phi_{k-1}(h))$.
Note that $(x+\phi_k(h))-(x+\phi_{k-1}(h)) = h_k e_k$, where $e_k$ is the standard $k$th unit vector $(\cdots, 0,1,0, \cdots)$.
Apply the mean value theorem to get $f(x+\phi_k(h))-f(x+\phi_{k-1}(h)) = {\partial f(x+\phi_{k-1}(h)+ \xi_k e_k) \over \partial x_k} h_k$, where $|\xi_k| \le |h_k|$.
Hence we have the estimate $f(x+h)-f(x) - \sum_k {\partial f(x) \over \partial x_k} h_k = \sum_k ({\partial f(x+\phi_{k-1}(h)+ \xi_k e_k) \over \partial x_k}-{\partial f(x) \over \partial x_k}) h_k$, again with $|\xi_k| \le |h_k|$.
Now choose $\epsilon>0$ and choose $\delta>0$ such that if the $|h_k| < \delta$, then $|{\partial f(x+\phi_{k-1}(h)+ \xi_k e_k) \over \partial x_k}-{\partial f(x) \over \partial x_k}| < {1 \over n} \epsilon$. We can do this because the partials are continuous at $x$.