$\displaystyle 3x^2y+2x^2-y^3-2y^2=\alpha{} \\[2ex] \textit{Differentiating this w.r.t x}\\[2ex] \displaystyle 3\left(2xy+x^2y^{'}\right)+4x-3y^2y^{'}-4y\ y^{'}=0 \\[2ex] \textit{To find the orthogonal trajectory, replacing }\\[2ex] \displaystyle y^{'}by-\frac{1}{y'} \\[2ex] \displaystyle 3\left(2xy-\frac{x^2}{y'}\right)+4x+\frac{3y^2}{y'}+\frac{4y}{y'}=0 \\[4ex] $

$\textit{Multiplying throughout by y'}\\[4ex] \displaystyle 3\left(2xyy^{'}-x^2\right)+4xy^{'}+3y^2+4y=0\ \\[2ex] \displaystyle 6xyy^{'}+4xy^{'}-3x^2+3y^2+4y=0 \\[2ex] \displaystyle \left(6xy+4x\right)\ \frac{dy}{dx}\ -3x^2+3y^2+4y=0 \\[2ex] \displaystyle \left(6xy+4x\right)\ dy+\left(-3x^2+3y^2+4y\right)dx=0 \\[2ex] $

$\textit{Hence the Differential equation is an exact equation. }\\[2ex] \begin{align*} \displaystyle f\left(x,\ y\right)&=\int{}M\ dx\\[2ex]&=\int{}\left(-3x^2+3y^2+4y\right)\ dx\\[2ex]& =\ -3\frac{x^3}{3}+3xy^2+4xy+f\left(y\right)\\[2ex]&=\ -x^3+3xy^2+4xy+f(y) \\[2ex] \end{align*} $

$\begin{align*} \displaystyle f\left(x,y\right)&=\int{}Ndy\\[2ex]&=\int{}\left(6xy+4x\right)dy\\[2ex]&=6x\frac{y^2}{2}+4xy+g\left(x\right)\\[2ex]&=3xy^2+4xy+g(x) \\[2ex] \end{align*}$

$ \textit{Comparing the above two, }\\[2ex] \displaystyle f\left(y\right)=0,\ g\left(x\right)=\ -x^3 \\[2ex] \textit{So, the general solution is}\\[2ex] \displaystyle \therefore{}3xy^2-x^3+4xy=\alpha{} \\[2ex] \textit{Which is the orthogonal trajectory to} \\[2ex]3x^2y+2x^2-y^3-2y^2=\alpha{}. $