Fake proof of differentiability It's a theorem that if $f\colon U\subset\Bbb R^n\to \Bbb R^m$ has the property that each of the partial derivatives $\partial_if_j$ exist and are continuous $p\in U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the  continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + \sum_i \partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| \to 0$ as $v\to 0$. Hence, we can write
\begin{align*}
f(p+v) &= f(p) + \big(\sum_i \partial_if_1(p)v_i + R_1(v),\dots,\sum_i \partial_if_m(p)v_i + R_m(v)\big) \\
&= f(p) + \sum_j\big(\sum_i\partial_if_j(p)v_i\big)e_j + R_j(v)e_j \\
&= f(p) + [Df_p][v] + (R_1,\dots,R_m)(v),
\end{align*}
where $[Df_p] = [\partial_if_j(p)]$ is the usual Jacobian  matrix, and $[v]$ is the column vector $[v_1\ \dotsb\ v_n]^T$. Now,
$$
\frac{|(R_1,\dots,R_m)(v)|^2}{|v|^2} = \frac{R_1(v)^2 + \dots + R_m(v)^2}{|v|^2} \to 0,
$$
where the last expression goes to $0$ as $v\to 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.

In short, if you don't assume that the $\partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
