= Solution
Let $F(y)=y_1^2y_2+y_2^2y_3+y_3^2y_1$. The <proper transform> is
$$
\boxed{\widetilde Y=\{(x,[y]):x_iy_j=x_jy_i\ (i<j),\ F(y)=0\}\subset\mathbb A^3\times\mathbb P^2.}
$$
Indeed, in the blowup chart $y_i\ne0$, write $x=t y$ with $y_i=1$. Homogeneity gives $F(x)=t^3F(y)$. 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 $(t y,[y])$, so no component in this displayed <set> is extraneous.
For $x\ne0$, 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 $2y_1y_2=-y_3^2$, $y_1^2=-2y_2y_3$, $y_2^2=-2y_3y_1$ gives $1=-8$, impossible. Thus \b[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.
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