Partial Derivatives Continuous does not guarantee Gradient Function continuous Just want to check my understanding that partial derivatives continuous does not mean that the gradient function $\nabla f$ is continuous. Is that correct?
E.g. $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ continuous does not necessarily mean that $\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ is continuous.
I concluded that based on "Continuity in each argument is not sufficient for multivariate continuity".
Question 2) What conditions guarantee that $\nabla f$ is continuous?
Thanks for any help.

We have that $\nabla f$ is continuous. Note that:
\begin{align*}
 || \nabla f(x,y) - \nabla f(x_0,y_0)||^2 = (\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0))^2 + (\frac{\partial f}{\partial y}(x,y)- \frac{\partial f}{\partial y}(x_0,y_0))^2
\end{align*}
Thus continuity of the partial derivatives imply that the gradient is continuous. 
Your quote says something different too. It says that continuity in x and y resp. does not imply continuity in $(x,y)\in \mathbb R^2$. This makes perfect sence, in $\mathbb R^2$ you can approach a point from many more directions that just parallel along the $x$- and $y$-axis.
DETAILS:
Let $\epsilon >0$ be fixed. Then by continuity of the partial derivatives we know there exist $\delta_1 >0$ and $\delta_2>0$ so that:
\begin{align*}
 ||(x,y)-(x_0,y_0)|| <\delta_1 \quad \Rightarrow \quad|\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0)| < \frac{\epsilon}{\sqrt{8}}\\
||(x,y)-(x_0,y_0)|| <\delta_2 \quad \Rightarrow \quad|\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0)| < \frac{\epsilon}{\sqrt{8}}
\end{align*}
Now let $\delta = \min \{\delta_1, \delta_2 \}$. Now we have that:
\begin{align*}
 || \nabla f(x,y) - \nabla f(x_0,y_0)||^2 &= (\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0))^2 + (\frac{\partial f}{\partial y}(x,y)- \frac{\partial f}{\partial y}(x_0,y_0))^2\\
 & < \frac{2\epsilon^2}{8} = \frac{\epsilon^2}{4}.
\end{align*}
Thus we have:
\begin{align*}
 || \nabla f(x,y) - \nabla f(x_0,y_0)|| < \frac{\epsilon}{2},
\end{align*}
as required.
