Directional derivatives at certain points 
Consider the function $f(x,y,z) = |x+y+z|$. If ${\bf a} =
 (a_1,a_2,a_3)$ satisfies $a_1+a_2+a_3=0$ find those directions ${\bf
 v}$ in which the derivative at ${\bf a}$ exists.

Attempt
Note that $f(a_1,a_2,a_3) = 0$. So, if ${\bf v } = (v_1,v_2,v_3)$ is any vector we have
$$ D_{\bf v} f( {\bf a} ) = \lim_{s \to 0} \frac{ f( a_i + s v_i ) - f(a_i ) }{s} $$
but $f(a_i) = 0$ and $f(a_i+sv_i) = |a_1 + sv_1 + a_2 + sv_2 + a_3 + sv_3| = s |v_1+v_2+v_3|$
So,
$$ D_{\bf v} f( {\bf a}) = \lim_{s \to 0} \frac{  s |v_1+v_2+v_3| }{s} = |v_1+v_2+v_3| = f ({\bf v})$$
but inst this hold for any $v$ and so the derivative of $f$ at ${\bf a}$ exists at all directions. Is this correct?

The directional derivative exists if limits as $s \to 0+$ from the right and as $s \to 0-$ from the left both exist and are equal.
You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| \neq s|v_1 + v_2 +v_3|$ if $s < 0$.
Thus,
$$\frac{f(\mathbf{a}+ s\mathbf{v}) - f(\mathbf{a})}{s} = \frac{|s|}{s}|v_1+v_2 + v_3| = \text{sgn}(s)|v_1 + v_2 + v_3|,$$
and left- and right-limits are unequal (due to opposite sign)  unless $|v_1 + v_2 +v_3| = 0$
