Solution: We have \begin{align*} & \left( x^3 + 3x^2 \right) \mathrm{d} x + \left( y^3 + 3x^{2} y \right) \mathrm{d} y = 0 \\ \implies & \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-x^3 - 3x y^2 }{y^3 + 3x^2 y} = - \frac{1 + 3 \left( \frac{y}{x} \right) }{\left( \frac{y}{x} \right) ^3 + 3 \left( \frac{y}{x} \right) }. \end{align*} Take $v = \frac{y}{x}$, which implies \[ y = vx \implies \frac{\mathrm{d} y}{\mathrm{d} x} = v + x \frac{\mathrm{d} v}{\mathrm{d} x}. \]

Therefore, \begin{align*} & v + x \frac{\mathrm{d} v}{\mathrm{d} x} = - \frac{1 + 3v^2 }{v^3 + 3v} \\ \implies & x \frac{\mathrm{d} v}{\mathrm{d} x} = - \frac{1+ 3v^2}{v^3 + 3v} - v \\ \implies &x \frac{\mathrm{d} v}{\mathrm{d} x} = - \frac{v^4 + 6v^2 + 1}{v^3 + 3v} \\ \implies & \frac{\mathrm{d} x}{x} = -\left( \frac{v^3 + 3v}{v^4 + 6v^2 + 1} \right) \mathrm{d} v \\ \implies & \int \frac{\mathrm{d} x}{x} = - \int \frac{v^3 + 3v}{v^4 + 6v^2 + 1} \mathrm{d} v \\ \implies & \ln x = -\frac{1}{4} \int \frac{\mathrm{d} t}{t},~\text{take } t= v^4 + 6v^2 + 1 \\ \implies & \ln x^4 = \ln \frac{c}{t},~ c \text{ is a constant} \\ \implies & \ln x^4 = \ln \left( \frac{c}{v^4 + v^2 _ 1} \right) \\ \implies & x^4 + y^4 + 6x^2 y^2 = c. \end{align*}