$2xy \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=-4{x}^{3}$
Separate the differentials$2xy\mathrm{d}^y=-4{x}^{3}\mathrm{d}^x$
Divide both sides of the equation by $x$$2y\mathrm{d}^y=\frac{ -4{x}^{3} }{ x }\mathrm{d}^x$
Integrate the left-hand side of the equation with respect to $y$ and the right-hand side of the equation with respect to $x$$\int{ 2y } \mathrm{d} y=\int{ \frac{ -4{x}^{3} }{ x } } \mathrm{d} x$
Evaluate the integral${y}^{2}+C_1=\int{ \frac{ -4{x}^{3} }{ x } } \mathrm{d} x$
Evaluate the integral$\begin{array} { l }{y}^{2}+C_1=-\frac{ 4{x}^{3} }{ 3 }+C_2,& \begin{array} { l }C_1 \in ℝ,& C_2 \in ℝ\end{array}\end{array}$
Since both constants of integration $C_1$ and $C_2$ are arbitrary constants, replace them with the constant $C$$\begin{array} { l }{y}^{2}=-\frac{ 4{x}^{3} }{ 3 }+C,& C \in ℝ\end{array}$