I don't want a full solution per se, just an appropriate hint/guidance to nudge me in the right direction if possible.
The problem at hand: $$(dy+dx)^3 = 27(x+y)^2(dx)^3$$
Thus far I've tried using a substitution $z=\frac{dy}{dx}$, and since $zdx=dy$ I'm able to simplify the expression to the following: $$(z+1)^3=27(x+y)^2\\$$ But this results in quite a hairy integral equation, dependent on both x and y, when I try solving for z, which leads me to believe that I am missing some detail (and gauging from previous experience, most likely a very obvious one).
Update: I thought of letting $v=x+y$. Thus $dv=1+\frac{dy}{dx}=z+1$ which has lead to finding one answer, that being $$y+x = (x+c)^3$$
$$(1+y')^3=(x+y)'^3=27(x+y)^2$$ Try to substitute $u=x+y$: $$u'^3=27u^2$$ Consider two cases $$u=x+y=0 \implies y=-x$$ This is a solution of the DE. And $u\ne 0$ gives another solution. $$\int \dfrac {du}{u^{2/3}}=3\int dx$$