We consider the blow up $Bl(\mathbb{P}^2)_p$ of $\mathbb{P}^2$ in $p:=|1:0:0|$ and the following surface:
$Y:=\{(|y_1: y_2:y_3:y_4|, |x_0:x_1:x_2|) \in \mathbb{P}^3\times \mathbb{P}^2: rk(\begin{pmatrix}y_1 & y_2 & y_3 \\ x_0 & x_1 &x_2\end{pmatrix}=1 , x_2^n=x_0^n+x_1^n\}$.
Then we can define the following map $\rho: Y\to Bl(\mathbb{P}^2)_p$:
$\rho(|y_1: y_2:y_3:y_4|, |x_0:x_1:x_2|)= (|y_4^n: y_1^n: y_2^n|,|x_0^n:x_1^n|)$
What is the branch locus?
I think it is
$branch(\rho)= l_1+l_2+l_3+l_4+E$
where $E$ is the exceptional locus of the blow-up and $l_1:=Z(a_0), l_2:=Z(a_1), l_3:=Z(a_0+a_1)$, and $l_4:=Z(t_0)$, where $(|t_0: t_1 :t_2|, |a_0:a_1|)$ are the coordinates of $Bl(\mathbb{P}^2)_p$
This map corresponds to the quotient map of $Y$ with respect the following action of $(\mathbb{Z}/n\mathbb{Z})^3$:
$(a_1,a_2,a_3)\cdot (y, x)=(|e^{\frac{2\pi i}{n}a_1}y_1: e^{\frac{2\pi i}{n}a_2}y_2:e^{\frac{2\pi i}{n}a_3}y_3:y_4|, |e^{\frac{2\pi i}{n}a_1}x_0:e^{\frac{2\pi i}{n}a_2}x_1: e^{\frac{2\pi i}{n}a_3}x_2|)$
