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).
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $$f_j$$ is a differentiable function on $$\mathbb R^{n}$$.