system of congruence - my approach We have:
$$k^3 + l^3 \equiv 0 \pmod{17}\\
k^2 + l^2 \equiv 0 \pmod{17} $$
And I get:
$$k = 17n+r_k\\
l = 17m+r_l$$
And I analyzed possible rests respect to system of congruences.
My result is: 
$$ k,l \equiv 0\pmod{17}$$
But is it properly ? Is my way is ok ? Is exists other, better solution ?

For any integer $a,a\equiv0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7,\pm8\pmod{17}$
$$\implies a^2\equiv0,1,4,9,16\equiv-1,8,2,15,13$$
Find which combinations of $(k,l)$ gives $k^2+l^2\equiv1\pmod{17}$ namely, obvious sets being $(0,\pm1);(\pm3,\pm3),(\pm4,\pm6)$
Find which of them, satisfy  $k^3+l^3\equiv1\pmod{17}$
Answer to the EDITED Question
Observe that  $17|k\iff 17|l$
Else we have $$(k\cdot l^{-1})^2\equiv-1\ \ \ \ (1),(k\cdot l^{-1})^3\equiv-1\pmod{17}\ \ \ \ (2)$$
$k\cdot l^{-1}\equiv\pm4,$ satisfy $(1)$ but not $(2)$
