Let $ f : \mathbb R^N \longrightarrow \mathbb R^N $ defined by
$$f(x) = |x|^{p-2}x$$
for all $ x \in \mathbb R^N $ with $p >1 $. Where $|x|$ is the euclidean norm in $ \mathbb R^N $.
I need to calculate the jacobian $J f(a)$ in any fixed point $ a \in \mathbb R^N $.
By definition we need to show that there is a linear transformation $ T_{a}: \mathbb R^N \longrightarrow \mathbb R^N $ such that
$$f(a + h) := f(a) + T_{a}(h) + r(h)$$
where $ \lim_{h \to 0}\frac{r(h)}{|h|} = 0 $.
Anyone have any ideas that can help me?
First, your intuition is wrong (and I really cannot imagine where you got that "intuition": in the case where $n=1$ and $x>0$, you know that $T_ah=(p-1)a^{p-2}h$). For $n>1$, it would be very unusual that $T_a$ is a scalar multiple of the identity, as you propose.
For $a\ne0$, the function $f$ is differentiable since $|x|$, exponentials, and products are differentiable.
For $a\ne0$: It is well known that the entries of $T_a$ in the canonical basis are $$ (T_a)_{kj}=\frac{\partial f_k}{\partial x_j}(a). $$ As $f_k(x)=|x|^{p-2}x_k=(x_1^2+\cdots+x_n^2)^{p/2-1}x_k$, \begin{align} (T_a)_{kj}&=\left.2(p/2-1)(x_1^2+\cdots+x_n^2)^{p/2-2}x_k^2+(x_1^2+\cdots+x_n^2)^{p/2-1}\right|_{x=a}\\ \ \\ &=|a|^{p-2}\,\left[(p-2)(a_1^2+\cdots+a_n^2)^{-1}a_k^2+1 \right]. \end{align}
Now for $a=0$.
For $p<2$: a necessary condition for differentiability at a point is that the partial derivatives exist at such point. We have, for $j\ne k$, $$ \frac{\partial f_k}{\partial x_j}(0)=\lim_{t\to0}\frac{f_k(te_j)-f_k(0)}{t} =\lim_{t\to0}\frac0t=0; $$ for $j=k$, $$ \frac{\partial f_k}{\partial x_k}(0)=\lim_{t\to0}\frac{f_k(te_k)-f_k(0)}{t} =\lim_{t\to0}\frac{t^{p-1}}t=\lim_{t\to0}t^{p-2}, $$ which doesn't exist. So $f$ is not differentiable at $0$.
For $p=2$, you have $f(x)=x$, which is obviously differentiable everywhere and $T_a=I$ for all $a$.
For $p>2$, the computations above show that $\partial f_k/\partial x_j(0)=0$. And one can check that $\lim_{a\to0}(T_a))_{kj}=0$, so the partial derivatives are continuous and so $f$ is differentiable at zero.