 Solution

Let $F(y)=y_1^2{y}_2+y_2^2{y}_3+y_3^2{y}_1$. The proper transform isIndeed, in the blowup chart $y_i\ne 0$, write $x=ty$ with $y_i=1$. Homogeneity gives $F(x)=t^{3}F(y)$. Away from the exceptional divisor $t=0$, the equation is $F(y)=0$, and its closure retains that equation, removing the exceptional factor $t^3$. Every point with $t=0,F(y)=0$ is approached algebraically along the line $(ty,[y])$, so no component in this displayed set is extraneous.

For $x\ne 0$, the fibre is the single point $[y]=[x]$, and the projection is an isomorphism over $Y\setminus \{0\}$. Over $0$ the fibre is the projective plane cubic $F(y)=0$. In characteristic zero this cubic is smooth: a coordinate zero in the three derivative equations would force all coordinates zero; with all nonzero, multiplying $2y_1{y}_2=-y_3^2$, $y_1^2=-2y_2{y}_3$, $y_2^2=-2y_3{y}_1$ gives $1=-8$, impossible. Thus the exceptional fibre is a smooth genus-one plane cubic. The fibre description itself does not require this optional smoothness conclusion; characteristic $3$ requires separate treatment.