= Solution
Homogenizing the affine equation gives the <projective closure>
$$
C=\{[X:Y:Z]\in\mathbb P^2:
X^3Y+Y^3Z+Z^3X=0\}.
$$
This is the <Klein quartic>. It is smooth: if all three projective partial derivatives vanished, then
$$
3X^2Y=-Z^3,
\qquad
3Y^2Z=-X^3,
\qquad
3Z^2X=-Y^3.
$$
Multiplying them gives $28X^3Y^3Z^3=0$. If one coordinate is zero, the displayed equations force all three to vanish, which is impossible in projective space. Thus $C$ is a smooth compactification of $X'$. The smooth compactification compatible with the extended coordinate functions is its normalization, and since $C$ is already smooth it is biholomorphic to $C$.
For a direct genus computation, consider the extended coordinate map $\pi_x:X\to\mathbb C_\infty$. A generic $x$ leaves the cubic equation
$$
y^3+x^3y+x=0,
$$
so $\deg\pi_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven points
$$
y^7=-\frac38,
\qquad
x=2y^3.
$$
At the origin, $x=-y^3+O(y^6)$, so the <ramification index of a holomorphic map> is three and the contribution is two. At each of the other seven points, $F_x=-7/2\ne0$, $F_y=0$, and $F_{yy}=6y\ne0$, so the ramification index is two and each contributes one.
There are two points at infinity. Near $P=[1:0:0]$, in coordinates $u=Y/X$, $v=Z/X$, the equation is
$$
u+u^3v+v^3=0,
$$
and $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation is
$$
r^3+s+rs^3=0.
$$
Thus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is therefore
$$
2+7+1=10.
$$
The <Riemann-Hurwitz formula> for the degree-three map to the sphere gives
$$
2g(X)-2=3(2\cdot0-2)+10=4,
$$
and hence
$$
\boxed{g(X)=3}.
$$
This also agrees with the <genus of a smooth plane curve> of degree four.
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