Derivative function continuous iff partial derivatives continuous 
Let $f:\mathbb{R} ^{n}\rightarrow \mathbb{R} ^{m}$ be differentiable.
The derivative function $Df:\mathbb{R} ^{n}\rightarrow L\left( \mathbb{R} ^{n},\mathbb{R} ^{m}\right)$ is continuous in respect to the operator norm $\left\| A \right\|_{L\left( \mathbb{R} ^{n},\mathbb{R} ^{m}\right)}:=\sup _{\left\| v\right\| =1}\left\| Av\right\|$, iff the partial derivatives $\dfrac {\partial f_{i}}{\partial x_{j}}$ are continuous for all $i\in \left\{ 1,\ldots ,m\right\}$ and $j\in \left\{ 1,\ldots ,n\right\}$.

How can I show this?

The idea behind all the proofs I've seen is to use the mean value theorem (or mean value inequality if you're working in general Banach spaces). This is carried out in a clear fashion in  Henri Cartan's book Differential calculus in proposition 3.7.2. BTW this book is out of print, but I think there is a reprint under a different name; see https://www.amazon.com/Differential-Calculus-Normed-Spaces-Analysis/dp/154874932X. There is also a proof in Loomis and Sternberg's book Advanced Calculus in Theorem 8.2 of Chapter 3. I HIGHLY recommend both these books. You can also find a proof in Spivak's Calculus on Manifolds, in Theorem 2-8 (Spivak only proves the "if" part).
The "only if" part is pretty much trivial once you know how $Df(a)$ and the various partials are related (see either Cartan/ Loomis and Sternberg).
As an outline for the "if" part, it suffices to prove it in the case $m=1$ (it's easy to deduce the general case from this). Notice the following equality:
\begin{align}
&  f(x_1, \dots, x_n) - f(a_1, \dots, a_n) - \sum_{i=1}^n \dfrac{\partial f}{\partial x_i}(a) \cdot (x_i-a_i) \\
&= f(x_1, x_2, \dots x_n) - f(a_1, x_2, \dots, x_n) - \dfrac{\partial f}{\partial x_1}(a) \cdot (x_1-a_1) \\\\
&+ f(a_1, x_2, \dots, x_n) - f(a_1, a_2, \dots, x_n) - \dfrac{\partial f}{\partial x_2}(a) \cdot (x_2-a_2) \\
& \vdots  \\
&+ f(a_1, \dots, a_{n-1}, x_n) - f(a_1, \dots, a_{n-1}, a_n) - \dfrac{\partial f}{\partial x_n}(a) \cdot (x_n-a_n)
\end{align} 
Now, applying the mean-value theorem (the standard single variable version) to each line separately, and using the continuity of the partials allows you to complete the proof.
