Homogenizing the affine equation gives the <projective closure>
$$
X=V(F)\subset\mathbb P^2,
\qquad
F(X,Y,Z)=Y^3Z-X^4-Z^4.
$$
Its partial derivatives are
$$
F_X=-4X^3,
\qquad
F_Y=3Y^2Z,
\qquad
F_Z=Y^3-4Z^3.
$$
If all three vanished at a point of $X$, then $X=0$ and either $Y=0$ or $Z=0$. If $Y=0$, the equation $F=0$ forces $Z=0$; if $Z=0$, the equation $F_Z=0$ forces $Y=0$. Either conclusion gives the forbidden zero triple of <homogeneous coordinates>. The <Jacobian criterion> therefore shows that $X$ is a <smooth projective curve>; more specifically, it is the <projective closure of y cubed equals x to the fourth plus one>, a <smooth plane quartic>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 24F / a / Solution

{title2=$Y^3Z=X^4+Z^4$}

The <projective closure> of the <superelliptic curve> $y^3=x^4+1$ is the smooth plane quartic
$$
Y^3Z=X^4+Z^4.
$$
Its unique point at infinity is $P_\infty=[0:1:0]$. The rational function $x=X/Z$ defines a degree-three morphism to $\mathbb P^1$ ramified with index three at $P_\infty$ and at the four affine points $(\alpha,0)$ with $\alpha^4=-1$. Its genus is three, and
$$
\operatorname{div}\left(\frac{dx}{y^2}\right)=4P_\infty.
$$

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