$\displaystyle dy/dx = -6xy$
$\displaystyle \int \frac{1}{y} dy = \int -6x dx$
$\displaystyle ln y = -3x^2 + c$
$\displaystyle y = e^{-3x^2 + c} $
is my answer Correct ?
First note that $\displaystyle \displaystyle \begin{align*} \int{\frac{1}{y}\,dy} = \ln{|y|} \end{align*}$, NOT $\displaystyle \displaystyle \begin{align*} \ln{(y)} \end{align*}$ so this means the correct working out would be
$\displaystyle \displaystyle \begin{align*} \ln{|y|} &= -3x^2 + C \\ |y| &= e^{-3x^2 + C} \\ |y| &= e^{C}\,e^{-3x^2} \\ y &= \pm e^{C}\,e^{-3x^2} \\ y &= A\,e^{-3x^2} \textrm{ where } A = \pm e^C \end{align*}$