\( x y ^ { 2 } d y - ( x ^ { 3 } + y ^ { 3 } ) d x = 0 \)

To solve the differential equation \[ x y^2 \, dy - (x^3 + y^3) \, dx = 0, \] we can rearrange it into a more standard form. Dividing through by \(dx\) gives: \[ x y^2 \frac{dy}{dx} = x^3 + y^3. \] This can be rewritten as: \[ \frac{dy}{dx} = \frac{x^3 + y^3}{x y^2}. \] Next, we can separate the variables. We can rewrite the equation as: \[ \frac{dy}{dx} = \frac{x^3}{x y^2} + \frac{y^3}{x y^2} = \frac{x^2}{y^2} + \frac{y}{x}. \] Now, we can separate the variables: \[ \frac{dy}{\frac{x^2}{y^2} + \frac{y}{x}} = dx. \] This is a bit complicated to integrate directly, so let's try a substitution. We can use the substitution \(v = \frac{y}{x}\), which implies \(y = vx\) and \(dy = v \, dx + x \, dv\). Substituting these into the equation gives: \[ v \, dx + x \, dv = \frac{x^3 + (vx)^3}{x (vx)^2}. \] This simplifies to: \[ v \, dx + x \, dv = \frac{x^3 + v^3 x^3}{x^3 v^2} = \frac{1 + v^3}{v^2}. \] Now, we can separate the variables: \[ x \, dv = \frac{1 + v^3}{v^2} \, dx - v \, dx. \] This leads to: \[ x \, dv = \left(\frac{1 + v^3 - v^2}{v^2}\right) dx. \] Now we can integrate both sides. The left side integrates with respect to \(v\) and the right side with respect to \(x\). However, this approach can become quite complex. Instead, let's return to the original equation and check if it is exact or can be made exact. The original equation can be written as: \[ M(x, y) = - (x^3 + y^3), \quad N(x, y) = x y^2. \] To check if it is exact, we compute: \[ \frac{\partial M}{\partial y} = -3y^2, \quad \frac{\partial N}{\partial x} = y^2. \] Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact. Next, we can try to find an integrating factor or use another method to solve the equation. After some manipulation, we can find that the equation can be solved using the method of integrating factors or substitutions, but it may require more advanced techniques or numerical methods depending on the context. For a specific solution, we can also look for particular solutions or numerical solutions based on initial conditions if provided. If you have specific conditions or further instructions, please let me know!