Imagine a landscape, where $x$ is west-east, and $y$ is north-south. Let $f(x,y)$ be the height at that point. If we take the gradient of $f$, we get a function from $\mathbb{R}^2$ to $\mathbb{R}^2$. At $(x,y)$, $\nabla f(x,y)$ points "uphill"; that is, it points in the direction of steepest ascent.
This is useful for physics. If you have some kind of potential energy function, taking the gradient will get you the corresponding force field. For example, consider a point mass $M$ fixed at the origin, and a smaller mass $m$ elsewhere. The potential energy of the small particle is $U = -G \frac{Mm}{r}$. If you take the gradient, you'll get $\vec{F} = G \frac{Mm}{r^2}$, which corresponds* with the results from physics.
We have to convert to rectangular coordinates, take the gradient, then convert back to polar.
$$
\begin{align*}
U &= -G \frac{Mm}{r} \\
&= -G \frac{Mm}{\sqrt{x^2 + y^2 + z^2}} \\
\frac{\partial U}{\partial x} &= G \frac{Mm}{2 (x^2 + y^2 + z^2)^{3/2}} \cdot 2x = G \frac{Mm \cdot x}{(x^2 + y^2 + z^2)^{3/2}} \\
\frac{\partial U}{\partial y} &= G \frac{Mm \cdot y}{(x^2 + y^2 + z^2)^{3/2}} \\
\frac{\partial U}{\partial z} &= G \frac{Mm \cdot z}{(x^2 + y^2 + z^2)^{3/2}} \\
\nabla U &= \left\langle \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right\rangle \\
&= G \frac{Mm}{(x^2 + y^2 + z^2)^{3/2}} \langle x, y, z \rangle \\
&= G \frac{Mm}{x^2 + y^2 + z^2} \frac{\langle x, y, z \rangle}{\sqrt{x^2 + y^2 + z^2}} \\
&= G \frac{Mm}{r^2} \hat{r}
\end{align*}
$$
*Really, we should take the negative of the gradient, because in physics, we want our force vectors pointing downhill, towards lower energy. But flipping the sign is easy.
The magnitude of the gradient represents how fast the function changes along the gradient. The gradient vector is the first term in a Taylor expansion of $f$, so if we work around a point $\vec {x_0}$ and consider a nearby point $\vec x$ we have $f(\vec x) \approx f(\vec{x_0}) + (\vec x - \vec {x_0})\cdot \vec{\nabla} f(\vec{x_0})$. The direction of maximum decrease is opposite to the gradient because it makes the cosine of the angle in the dot product $-1$. There is no information here about how far you should go in that direction. That would come from a second derivative.
Best Answer
Use the definition. If $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n x_k^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n x_k^2 ,$$ then $$\frac{\partial}{\partial x_j}f(x) =\frac{\partial}{\partial x_j}\sum_{k=1}^n x_k^2=\sum_{k=1}^n \underbrace{\frac{\partial}{\partial x_j}x_k^2}_{\substack{=0, \ \text{ if } j \neq k,\\=2x_j, \ \text{ else }}}= 2x_j.$$ It follows that $$\nabla f(x) = 2x.$$