Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.

Suppose for the sake of contradiction that $X$ is $k$-gonal for $k \leq d-1$. Let $(x_1, x_2, \ldots, x_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x_i$. (If our map is separable, then a generic fiber has distinct points. In general, our map is the composition of a power of Frobenius and a separable map, so we can reduce to the separable case.) Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x_1)$, $\phi(x_2)$, ..., $\phi(x_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.

Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x_i = (u_i: v_i : w_i)$. By genericity, we can assume that the $(u_i:v_i)$ are distinct. Then $d-2$ of the coordinates of $\phi(x_i)$ are $(u_i^{d-3}: u_i^{d-4} v_i : \cdots : v_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.

If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. But we will show that, in this case, the points $x_i$ are forced to be on a line.

In this case, we are requiring that the whole $(d-1) \times \binom{d-1}{2}$ matrix has rank $(d-2)$. The $(d-2)$ columns mentioned above already have rank $d-2$, since they are Vandermonde, so every other column is in their linear span. In particular, for every $j$, there is a degree $d-3$ polynomial $F_j(u,v)$ such that $F_j(u_i,v_i) = u_i^j v_i^{d-4-j} w_i$ for all $i$.

Then $u_i F_j(u_i, v_i) = v_i F_{j+1} (u_i, v_i)$. But $u F_j - v F_{j+1}$ is a degree $d-2$ homogenous polynomial so, if it vanishes at $d-1$ points, it is zero. We deduce that $u F_j = v F_{j+1}$ for $1 \leq j \leq d-5$, and thus that there is a degree $1$ polynomial $G$ such that $F_j(u,v) = u^j v^{d-4-j} G(u,v)$. Thus $w_i = G(u_i, v_i)$ for all $i$, and we have shown that the points lie on a line.

Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.

Suppose that $X$ is $k$-gonal for $k \leq d-1$. Let $(x_1, x_2, \ldots, x_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x_i$. (If our map is separable, then a generic fiber has distinct points. In general, our map is the composition of a power of Frobenius and a separable map, so we can reduce to the separable case.) Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x_1)$, $\phi(x_2)$, ..., $\phi(x_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.

Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x_i = (u_i: v_i : w_i)$. By genericity, we can assume that the $(u_i:v_i)$ are distinct and all $u_i$ and $v_i$ are nonzero. Then $d-2$ of the coordinates of $\phi(x_i)$ are $(u_i^{d-3}: u_i^{d-4} v_i : \cdots : v_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.

If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. But we will show that, in this case, the points $x_i$ are forced to be on a line.

In this case, we are requiring that the whole $(d-1) \times \binom{d-1}{2}$ matrix has rank $(d-2)$. The $(d-2)$ columns mentioned above already have rank $d-2$, since they are Vandermonde, so every other column is in their linear span. In particular, for every $j$, there is a degree $d-3$ polynomial $F_j(u,v)$ such that $F_j(u_i,v_i) = u_i^j v_i^{d-4-j} w_i$ for all $i$.

Then $u_i F_j(u_i, v_i) = v_i F_{j+1} (u_i, v_i)$. But $u F_j - v F_{j+1}$ is a degree $d-2$ homogenous polynomial so, if it vanishes at $d-1$ points, it is zero. We deduce that $u F_j = v F_{j+1}$ for $1 \leq j \leq d-5$, and thus that there is a degree $1$ polynomial $G$ such that $F_j(u,v) = u^j v^{d-4-j} G(u,v)$. Thus $w_i = G(u_i, v_i)$ for all $i$, and we have shown that the points lie on a line.

Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.

Suppose for the sake of contradiction that $X$ is $k$-gonal for $k \leq d-1$. Let $(x_1, x_2, \ldots, x_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x_i$. Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x_1)$, $\phi(x_2)$, ..., $\phi(x_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.

Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x_i = (u_i: v_i : w_i)$. By genericity, we can assume that the $(u_i:v_i)$ are distinct. Then $d-2$ of the coordinates of $\phi(x_i)$ are $(u_i^{d-3}: u_i^{d-4} v_i : \cdots : v_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.

If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. Even so, though, the full $(d-1) \times \binom{d-1}{2}$ matrix can only be rank $d-2$ if $w_i$ is a linear function of $(u_i:v_i)$ -- i.e. if the points are on a line.

Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.

Suppose for the sake of contradiction that $X$ is $k$-gonal for $k \leq d-1$. Let $(x_1, x_2, \ldots, x_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x_i$. (If our map is separable, then a generic fiber has distinct points. In general, our map is the composition of a power of Frobenius and a separable map, so we can reduce to the separable case.) Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x_1)$, $\phi(x_2)$, ..., $\phi(x_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.

Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x_i = (u_i: v_i : w_i)$. By genericity, we can assume that the $(u_i:v_i)$ are distinct. Then $d-2$ of the coordinates of $\phi(x_i)$ are $(u_i^{d-3}: u_i^{d-4} v_i : \cdots : v_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.

If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. But we will show that, in this case, the points $x_i$ are forced to be on a line.

In this case, we are requiring that the whole $(d-1) \times \binom{d-1}{2}$ matrix has rank $(d-2)$. The $(d-2)$ columns mentioned above already have rank $d-2$, since they are Vandermonde, so every other column is in their linear span. In particular, for every $j$, there is a degree $d-3$ polynomial $F_j(u,v)$ such that $F_j(u_i,v_i) = u_i^j v_i^{d-4-j} w_i$ for all $i$.

Then $u_i F_j(u_i, v_i) = v_i F_{j+1} (u_i, v_i)$. But $u F_j - v F_{j+1}$ is a degree $d-2$ homogenous polynomial so, if it vanishes at $d-1$ points, it is zero. We deduce that $u F_j = v F_{j+1}$ for $1 \leq j \leq d-5$, and thus that there is a degree $1$ polynomial $G$ such that $F_j(u,v) = u^j v^{d-4-j} G(u,v)$. Thus $w_i = G(u_i, v_i)$ for all $i$, and we have shown that the points lie on a line.