Answer

Taking the reciprocal on both sides,

\[ \begin{aligned} \frac{dx}{dy} &= 6e^{y} - 2x\newline \implies \diffone{x} + 2x &= 6e^{y}\newline \end{aligned} \]

which is a linear differential equation. Considering the integrating factor as \(exp \roundbr{\int 2dy} = exp \roundbr{2y}\),

\[ \begin{aligned} \implies \diffone{x}e^{2y} + 2xe^{2y} &= 6e^{3y}\newline \frac{d}{dy} \roundbr{xe^{2y}} &= 6e^{3y}\newline xe^{2y} &= \int 6e^{3y}dy + c = 2e^{3y} + c \end{aligned} \]

Another problem leveraging is the same transformation is

\[ \begin{aligned} \roundbr{y^{2} + 2x}\frac{dy}{dx} = y \end{aligned} \]