= Solution
For a <smooth projective curve> of <genus> $g$, the <Riemann-Roch theorem> states
$$
\ell(D)-\ell(K-D)=\deg D+1-g,
$$
where $D$ is a <divisor>, $K$ a <canonical divisor>, and $\ell(D)=\dim H^0(V,\mathcal O_V(D))$. In particular, applying it to $D=0$ and $D=K$ gives $\ell(K)=g$ and $\deg K=2g-2$.
For a nonconstant morphism $f:C\to D$ of degree $d$ in characteristic zero, choose local parameters with $t\circ f=u^e$ up to a unit at a point of <ramification index> $e$. Pulling back a differential multiplies its local order by $e$ and adds $e-1$, because $d(u^e)=eu^{e-1}du$ and $e\ne0$ in the field. Thus
$$
K_C\sim f^*K_D+\sum_{P\in C}(e_P-1)P.
$$
The <degree of a divisor> of a pulled-back <divisor> is $d$ times its degree. Taking degrees and using the <Riemann-Roch theorem> gives the <Riemann-Hurwitz formula>
$$
\boxed{2g_C-2=d(2g_D-2)+\sum_P(e_P-1).}
$$
A smooth plane cubic has <genus> one by the <genus-degree formula>. In the displayed normal form, $F(0,X_1)=cX_1^3$ with $c\ne0$: otherwise $X_0$ divides $F$ and the cubic is reducible. Therefore the line $X_0=0$ meets the cubic in the <divisor> $3P_0$, and $\mathcal O_V(1)\cong\mathcal O_V(3P_0)$. The <elliptic curve> group law identifies $P$ with the degree-zero <divisor> class $[P-P_0]$. Order three means $3P\sim3P_0$. Consequently $\mathcal O_V(3P)\cong\mathcal O_V(1)$ and has a section with zero <divisor> $3P$.
Every section of $\mathcal O_V(1)$ is the restriction of a linear form: ambient linear forms form a three-dimensional space and restrict injectively, while <Riemann-Roch theorem> gives $h^0(\mathcal O_V(1))=3$, since $\deg(K-3P_0)=-3$ forces the dual term to vanish. Thus this section is some nonzero $H$, proving \b[$V\cap V(H)=\{P\}$, with intersection multiplicity three]. This is the <three-torsion points are flexes of a plane cubic> property.
Projection forgetting $X_3$ defines a morphism $W\to V$: its only possible indeterminacy point $(0:0:0:1)$ does not satisfy the second equation. It is a double cover. Away from the zeros of $H_1H_2$, its two square roots are distinct. Each line intersects $V$ in three simple points, since it is nowhere tangent; the two triples are disjoint because the lines do not meet on $V$. In a local trivialization near any of these six points the covering equation is $s^2=u\,a(u)$ with $a(0)\ne0$, so the <ramification index> is two. There is no other ramification. The <Riemann-Hurwitz formula> therefore gives
$$
2g_W-2=2(2\cdot1-2)+6=6,\qquad\boxed{g_W=4.}
$$
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