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find an integrating factor and solve the given equation.
\begin{flalign*}
&(3x + \frac{6}{y} ) + (\frac{x^2}{y} + 3\frac{y}{x}) \frac{dy}{dx} = 0&\\
\end{flalign*}
simplify the equation and we will get
\begin{flalign*}
&(3x^2y + 6x)dx +(x^3 + 3y^2)dy = 0&\\
&M_y = 3x^2&\\
&N_x =3x^2&\\
&M_y = N_x&\\
\end{flalign*}
so it is exact\\
then find $\mu$
\begin{flalign*}
&\mu= \int(3x^2y+6x)dx&\\
&\mu = x^3y +3x^2 + h(y)&\\
&\mu _y= x^3 + h'(y)&\\
&\mu _y= N&\\
&h'(y) = 3y^2&\\
&h(y) = y^3 + c&\\
&\mu = x^3y + 3x^2 + y^3 +c&\\
&x^3y + 3x^2 + y^2 = c&
\end{flalign*}
solution is $x^3y + 3x^2 + y^2 = c$
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