Find all values $p$, $q$ for which $(\sum_{i=1}^n |x_i|^p )^{1/q}$ is differentable 
Find all values $p$, $q$ for which $(\sum_{i=1}^n |x_i|^p )^{1/q}$ is differentable in  $0 \in \mathbb R^n$

My observation:
$$\frac{\delta f}{\delta x_i}(x_1,...,x_n) =  \frac{p \left| x_i\right| ^{p-1} |x_i|' \left(\left| x_1\right| ^p+\left| x_2\right|^p+...+\left| x_{n-1}\right| ^p+\left| x_n\right| ^p\right)^{\frac{1}{q}-1}}{q} $$
but $ |x_i|' $ doesn't exists so there is no $p,q$ such  $(\sum_{i=1}^n |x_i|^p )^{1/q}$ is differentable
 A: First off, there's an issue with the case when $p,q<0$ (if only one of them is $<0$, the function diverges at $0$ anyway). Strictly speaking, in this case $f$ isn't defined if any $x_i=0$, but at the same time we can extend $f$ to these points by noting that if $a$ is any point which has one coordinate $0$, we have
$$
\lim_{x\rightarrow a}f(x)=0
$$
So we can just continuously extend $f$ to those points.  
Second, $q\neq0$ but if $p=0$ then we just have $f(x)=1$, which is differentiable.  
Otherwise, let's look at the directional derivative along $s=(1,...,1)$. By definition:
$$
\partial_sf(0)=\lim_{h\rightarrow 0}\frac{f(x_1+h,...,x_n+h)-f(0)}{h}
$$
In your case, this gives
$$
\partial_sf(0)=\lim_{h\rightarrow 0}\frac{(n|h|)^{p/q}}{h}=\lim_{h\rightarrow 0}n^{p/q}|h^{p/q-1}|\cdot\mathrm{sign}(h)
$$
This limit clearly only exists if $p/q-1>0$. If it does exist, it is $0$.  
So suppose $p/q-1>0$. Then to see that $f$ is differentiable, we need to evaluate the limit
$$
\lim_{h\rightarrow0}\frac{|f(0+h)-f(0)-\vec{0}\cdot h|}{\Vert h\Vert_2}=\lim_{h\rightarrow0}\frac{|f(h)|}{\Vert h\Vert_2}
$$
Here $h$ is now a vector and $\vec{0}$ is the $0$-vector. $\Vert\cdot\Vert_2$ is the standard norm.
Since we have the inequality 
$$f(h)=\left(\sum_{i=1}^n|h_i|^p\right)^{1/q}\leq\left(\sum_{i=1}^n\Vert h\Vert_2^p\right)^{1/q}=n^{p/q}\Vert h\Vert_2^{p/q}$$
and since we're dealing with the case $p/q-1>0$, the above limit is $0$ and the gradient is $\vec{0}$. (This also agrees with $\partial_s f(0)=0$, as it should.)
This means $f$ is differentiable at $0$ precisely when $p/q-1>0$ or $p=0$ (so long as we suitably extend $f$ in cases when $p,q<0$ - if not, then when $p\neq 0$ we have the additional condition that $p,q>0$).
