Multivariable calculus - find total derivative I want to find the total derivative of the function $f: \mathbb R^n \to \mathbb R^n$, $f(x)=\frac{x}{|x|}$

If I was to copy what the teacher taught, I should find the limit of $\lim_{t \to 0} \frac{f(a+th)-f(a)}{t}$ but I don't know how to find that limit.

$$\lim_{t \to 0} \frac{f(a+th)-f(a)}{t}=\lim_{t \to 0} \frac{\frac{a+th}{|a+th|}-\frac{a}{|a|}}{t}$$

what now?
For all $x\neq 0\in\mathbb R^n$ the function $f$ is differentiable at  $x$, i.e.
$$f(x+th)-f(x)=\langle \nabla f(x),th\rangle+O(t^2), $$
for all $t\in \mathbb R$ and $h\in \mathbb R^n$ s.t. $x+th\neq 0$, as the partial derivatives $\partial_i f(x)$ exist and are continuous in suitable open neighborhoods $U$ of $x$; this is show by   their explicit formulae, i.e.
$$\partial_i f(x):=\frac{\partial f}{\partial x_i}(x)=\frac{e_i}{\|x\|}-\frac{x_i}{\|x\|^3}x. $$
We denote by $e_i=(0,\dots,0,i,0,\dots,0)$ the $i$-th versor in $\mathbb R^n$.

Then
$$f(x+th)-f(x)=\frac{t}{\|x\|}\langle e_i,h\rangle-
\frac{t x_i}{\|x\|^3}\langle x,h\rangle +O(t^2), $$
which, in the limit $t\rightarrow 0$ gives

$$D_hf(x):=\lim_{t\rightarrow 0}\frac{f(x+th)-f(x)}{t}=\frac{h_i}{\|x\|}-
\frac{x_i\langle x,h\rangle}{\|x\|^3}. $$
Hint: I find it easier to break up problems that have absolute values in them into positive and negatives parts:
$$
  f(x) = \begin{cases}
          \cfrac{x}{x} &amp; \quad \text{for} \quad x \ge 0 \\
          \cfrac{x}{-x} &amp; \quad \text{for} \quad x &lt; 0
         \end{cases}
$$
For n-dimensional vectors you have to define what the norm means.  If it is a $L_2$ norm,
$$
  f(\mathbf{x}) = \frac{\mathbf{x}}{\sqrt{\mathbf{x}\cdot\mathbf{x}}}
$$
The derivative is directional and is defined in the direction $\mathbf{y}$ as
$$
  \frac{\partial f}{\partial \mathbf{x}}\cdot\mathbf{y} = \left[\frac{d}{d\alpha} f(\mathbf{x} + \alpha \mathbf{y})\right]_{\alpha = 0}
$$
Update: For the purposes of calculation it is easier if we move into index notation
$$
  \frac{\partial f}{\partial \mathbf{x}} \equiv \frac{\partial}{\partial x_j}\left(\frac{x_i}{\sqrt{x_k\,x_k}}\right)
$$
After working out the algebra we get
$$
  \frac{\partial f}{\partial \mathbf{x}} = \frac{\mathbf{I}}{||\mathbf{x}||} - \frac{\mathbf{x}\otimes\mathbf{x}}{||\mathbf{x}||^3} \,.
$$
Of course, these are all partial derivatives but, presumably, the total derivative can be found if these partial derivatives are known.