Find DE given its solution

Question

If anyone could give me a hint from where to start because I can't think by reversing solution...

$$ x^{3} y^{3} = 2^x - 6x + C \qquad\text{ }\quad $$

Answer 1

Since you have edited the tag 'differential-equations' I am going that way.(Well ..I don't have any other solution :P)

$$d(x^3y^3-2^x+6x)=d(C) $$

$$d(x^3y^3)-d(2^x)+d(6x)=0$$

$$3x^2y^3 ~dx+3y^2x^3~ dy-2^x \ln 2~dx+6~dx=0$$

$$-3y^2x^3~ dy=(3x^2y^3-2^x \ln 2+6)~dx$$

$$\implies \frac{dy}{dx}=\frac{3x^2y^3-2^x \ln 2+6}{-3y^2x^3}$$