= Solution
Let $H$ denote a <hyperplane-section divisor of a projective plane curve>. The <adjunction formula> for a smooth degree-$d$ <projective plane curve> states
$$
K_C\sim(d-3)H|_C.
$$
Since $\deg(H|_C)=d$, this gives the <canonical degree of a smooth plane curve>
$$
\boxed{\deg K_C=d(d-3)}.
$$
Combining this with $\deg K_C=2g-2$ gives the <genus of a smooth plane curve>
$$
\boxed{g(C)=\frac{(d-1)(d-2)}2}.
$$
<Homogenization> of the affine equation gives the <projective completion>
$$
C=V(F)\subset\mathbb P^2,
\qquad
F(X,Y,Z)=X^3Z+XY^3+YZ^3.
$$
This is the <plane model y plus x cubed plus xy cubed equals zero of the Klein quartic>: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^3B+B^3C+C^3A=0$.
Its first partial derivatives are
$$
F_X=3X^2Z+Y^3,
\qquad
F_Y=3XY^2+Z^3,
\qquad
F_Z=X^3+3YZ^2.
$$
If one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in <projective space>. If $XYZ\ne0$, multiplying the three derivative equations gives
$$
X^3Y^3Z^3=-27X^3Y^3Z^3,
$$
again a contradiction. The <Jacobian criterion> therefore proves that \b[$C$ is smooth].
The rational function $y=Y/Z$ defines a <rational map of projective varieties> $C\dashrightarrow\mathbb P^1$. Because $C$ is a <smooth projective curve>, it extends uniquely to a morphism
$$
\pi:C\longrightarrow\mathbb P^1.
$$
For a generic finite value of $y$, its fibre is given by the cubic
$$
x^3+y^3x+y=0,
$$
so the <degree of a holomorphic map> is $3$. The <discriminant of a depressed cubic> is
$$
\Delta(y)=-4y^9-27y^2=-y^2(4y^7+27).
$$
At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the <holomorphic implicit function theorem> gives
$$
y=-x^3+O(x^{10}),
$$
so $x$ is a <local coordinate> and $\pi$ has <ramification index of a holomorphic map> $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.
It remains to inspect the points at infinity. They are
$$
P=[1:0:0],
\qquad
Q=[0:1:0].
$$
Near $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^3v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a <simple pole>. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+st^3=0$, so $t=-s^3+O(s^{10})$ and
$$
y=\frac{s}{t}=-s^{-2}(1+O(s^7)).
$$
Thus $y$ has a <double pole> at $P$ and $e_P=2$. These calculations are the <ramification of the x-coordinate on the Klein quartic> after the coordinate relabelling above.
The total ramification contribution is
$$
(3-1)+7(2-1)+(2-1)=10.
$$
The <Riemann-Hurwitz formula> for the degree-three map to the <projective line> now gives
$$
2g(C)-2=3(-2)+10=4,
$$
and therefore
$$
\boxed{g(C)=3}.
$$
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