Question

dy/dx= −(x^2/y^2)+1/[3y^2(x3+ y3)^2]

Answer 1

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x^2}{y^2} + \frac{1}{3y^2(x^3 + y^3)^2}\\ 3y^2(x^3 + y^3)^2\frac{\mathrm{d}y}{\mathrm{d}x} = -3x^2(x^3 + y^3)^2 + 1 \\ 3y^2(x^3 + y^3)^2\frac{\mathrm{d}y}{\mathrm{d}x} + 3x^2(x^3 + y^3)^2 = 1\\ 3(x^3 + y^3)^2\left(y^2\frac{\mathrm{d}y}{\mathrm{d}x} + x^2\right) = 1\\ 3(x^3 + y^3)^2\left(y^2\frac{\mathrm{d}y}{\mathrm{d}x} + x^2\right) = 1\\ \frac{1}{3}\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3 + y^3\right)^3 = 3\\ \left(x^3 + y^3\right)^3 = \int \mathrm{d}x = 3(x + C)\\ x^3 + y^3 = \sqrt[3]{3(x + C)}\\ y^3 = \sqrt[3]{3(x + C)} - x^3\\ y= \sqrt[3]{\sqrt[3]{3(x + C)} - x^3}$