= Solution
The formula $[X:Y:Z]\mapsto[X:Z]$ is initially undefined at the unique point $P_\infty=[0:1:0]$ of $X$ at infinity. Since $X$ is a <smooth projective curve>, the <extension of a rational map from a smooth projective curve> extends it uniquely to a morphism
$$
\varphi:X\longrightarrow\mathbb P^1.
$$
On the affine chart it is the rational function $x$, and the equation $y^3=x^4+1$ shows that $\varphi$ has degree three.
For each of the four roots $\alpha$ of $x^4+1$, let $P_\alpha=(\alpha,0)$. Since the root is simple, $t=y$ is a <uniformizer> at $P_\alpha$, and the equation gives
$$
x-\alpha=c_\alpha t^3+O(t^6),
\qquad c_\alpha\ne0.
$$
Thus the <ramification index of a holomorphic map> at every $P_\alpha$ is three.
At $P_\infty$, use the chart $Y=1$ and put $u=X/Y$, $v=Z/Y$. The equation becomes
$$
v=u^4+v^4.
$$
The <implicit function theorem> makes $u$ a uniformizer and gives $v=u^4$ times a local unit. Hence
$$
x=\frac XZ=\frac uv=u^{-3}\mathbin{\cdot}(\text{local unit}),
$$
so $P_\infty$ also has ramification index three. These are all the <ramification points of a holomorphic map> because the resulting ramification degree is $5(3-1)=10$. Indeed, the <Riemann-Hurwitz formula> gives
$$
2g_X-2=3(2g_{\mathbb P^1}-2)+10=-6+10=4.
$$
Therefore $g_X=3$, in agreement with the <genus of a smooth plane curve>, and the <Euler characteristic> of the underlying compact orientable surface is
$$
\boxed{g_X=3,\qquad \chi(X)=2-2g_X=-4.}
$$
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