Why is Klein's quartic curve not hyperelliptic

Let $X$ be Klein’s quartic curve given by $x^3y + y^3z+z^3x=0$ in $\mathbf{P}^2$. It is isomorphic to $X(7)$.

How do I easily show that $X$ is not hyperelliptic?

I can see that $X$ is of genus $3$ and has gonality $\leq 3$ (consider the projection). I’m trying to prove that it has gonality $3$.

More generally, what is a computationally feasible way to check if a curve is not hyperelliptic?

Note that I’m not really asking for a criterion. For example, to check if a variety is normal you could try to show that it is regular (which is easier to me).

Is the obvious morphism $X\to \mathbf{P}^1$ of degree $3$ Galois? That is, do we have that $X$ is a cyclic cover of degree $3$?

Solutions Collecting From Web of "Why is Klein's quartic curve not hyperelliptic"

For a smooth curve $X$ of genus $g\geq 3$ (like the Klein quartic, which has genus $g=3$ as you remarked), the criterion you want is (Miranda , Chap.VII, Prop. 2.1):

$X$ is not hyperelliptic $\iff$ the canonical map $X\to \mathbb P^{g-1}$ is an embedding.

Conclude by remembering that if $X\subset \mathbb P^{g-1} $ is already embedded as a curve of of degree $2g-2$ (but not included in a hyperplane), then the canonical map is an embedding: Griffiths-Harris, page 247.