= Solution
A rational map $X\dashrightarrow Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.
For
$$
\phi(x:y:z)=(xy:xz:z^2),
$$
the coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.
Define
$$
\psi(u:v:w)=(v^2:uw:vw).
$$
Where all coordinates are nonzero,
$$
\psi\phi(x:y:z)=(x^2z^2:xyz^2:xz^3)=(x:y:z),
$$
and similarly $\phi\psi$ is the identity. Thus $\phi$ is birational and is an isomorphism on $\mathbb P^2\setminus Z(xyz)$.
For irreducibility of $P=x^2z^4-x^3y^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,
$$
y^3-\frac{x^2+1}{x^3}.
$$
The rational <function> on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.
Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation into
$$
C:\quad v^2w-u^3+w^3=0.
$$
Its <partial derivatives> are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.
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