Differentiability at $x_0 \in E$ implies continuity at $x_0$. We don't need continuity of the derivative or partial derivatives.
Indeed differentiability means that there exists a linear map $Df_{x_0}:\mathbb{R}^n \to \mathbb{R}^m$ such that for all $\epsilon > 0$ we have $$\|f(x) - f(x_0) - Df_{x_0}(x-x_0)\| < \epsilon \| x - x_0 \|$$
whenever $x$ is close enough to $x_0$. Fix any $\epsilon > 0$.
By the reverse triangle inequality in particular this means that
$$\|f(x) - f(x_0)\| - \|Df_{x_0}(x-x_0)\| < \epsilon \| x - x_0 \|,$$ or
$$\|f(x) - f(x_0)\| < \epsilon\|x-x_0\| + \|Df_{x_0}(x-x_0)\|.$$
But as $Df_{x_0}$ is linear and has the finite-dimensional domain $\mathbb{R}^n$, it is bounded, i.e., there is some constant $L > 0$ such that $\|Df_{x_0}(x-x_0)\| \leq L\|x-x_0\|$ for all $x$. Thus
$$\|f(x) - f(x_0)\| < (\epsilon+L)\|x-x_0\|.$$
Since we can make the right-hand side arbitrarily small by taking $x$ close enough to $x_0$, $f$ is continuous at $x_0$.