= Solution
The <genus-degree formula> gives <arithmetic genus> three for a <plane quartic>. Each of its three double points has <multiplicity of a plane curve at a point> two, so the <arithmetic genus drop under a point blowup> is one at each point. The resulting <integral projective curve> $C'$ has
$$
p_a(C')=3-3=0.
$$
For its finite <normalization of an algebraic curve> $\nu:\widetilde C\to C'$, the <arithmetic genus> identity is
$$
p_a(C')=g(\widetilde C)+\operatorname{length}(\nu_*\mathcal O_{\widetilde C}/\mathcal O_{C'}).
$$
Both terms are nonnegative, so $g(\widetilde C)=0$. Over the <algebraically closed field>, the <rationality of a smooth projective genus-zero curve> identifies $\widetilde C$ with the <projective line>: a point $P$ exists, and <Riemann-Roch theorem> gives a degree-one map from the two-dimensional space $L(P)$. Consequently $C$ is <birational> to $\mathbb P^1$. This proves the <rationality of a plane quartic with three double points> even when a double point is not an ordinary node.
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