(Newton 1671, “Problema II, Solutio particulare”). Solve the total differential equation
$$
3x^2-2ax+ay-3y^2y'+axy'=0.
$$
Solve:We have
$$
y'(3y^2-ax)=3x^2-2ax+ay.
$$
So
$$
dy(3y^2-ax)=(3x^2-2ax+ay)dx.
$$
So
$$
y^{3}-axy=x^3-ax^2+axy+C
$$
where $c$ is a constant.
Note that this solution is wrong,because $x$ and $y$ are dependent.