Klein quartic smooth in Char different from 7 I cannot solve the following exercise:
Let $\mathbb F$ be a field with $char(\mathbb F)\ne 7$. Let $C$ be the projective Klein curve defined by 
$x^3y+y^3z+z^3x =0$.
a) Prove that $C$ s smooth
b) Compute the divisors of the functions $x/y$ and $x/z$.
How can I prove that this curve is smooth only in $char(\mathbb F)\ne 7$?

The polynomial is fixed under the cyclic permutation of variables: $x\mapsto y\mapsto z\mapsto x$, so it suffices to study a single affine chart. I picked the chart with $z=1$.
At a singular point the following three equations must all hold simultaneously
$$
\left\{\begin{array}{lcl}
f=x^3y+y^3+x&=&0,\\
f_x=3x^2y+1&=&0,\\
f_y=x^3+3y^2&=&0.\end{array}\right.
$$
From the second equation we can solve $y=-1/3x^2$. Plugging this into the first gives
$$
\frac{2x}3-\frac1{27x^6}=0,
$$
implying that at a singularity $x^7=1/18$. OTOH, plugging it into the last equation gives
$$
x^3+\frac1{3x^4}=0
$$
Implying that $x^7=-1/3$. So for a singularity to exist we must have $18=-3$.
This holds in characteristics $3$ and $7$ only.


*

*In characteristic $3$ the second equation reads $1=0$ which is absurd.

*In characteristic $7$ we have the singular point $x=2,y=4$. Because there are no seventh roots of unity $x=2$ is the only solution, and this is therefore the only singular point. It may be worth pointing out that $2$ is the primitive third root of unity $\omega$ in characteristic seven, so the singular point can be rewritten in the homogeneous coordinates as
$$[x:y:z]=[1:\omega:\omega^2]$$ showing that this point really is stable under the cyclic permutation of coordinates.

