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Homogenizing the affine equation gives the projective closureThis is the Klein quartic. It is smooth: if all three projective partial derivatives vanished, thenMultiplying them gives $28X^3{Y}^3{Z}^3=0$. 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 $X^{\prime }$. 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 equationso $\mathrm{deg}{\pi }_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven pointsAt 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, $F_x=-7/2\ne 0$, $F_y=0$, and $F_{yy}=6y\ne 0$, 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 isand $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation isThus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is thereforeThe Riemann-Hurwitz formula for the degree-three map to the sphere givesand henceThis also agrees with the genus of a smooth plane curve of degree four.