Nonlinear PDE $u_y=(u_x)^3$

The Lagrange-Charpit equations tell us that:

$$\frac{\mathrm{d}x }{3p^2} = \frac{\mathrm{d} y}{-1} = \frac{\mathrm{d}u }{3p^3-q} = -\frac{\mathrm{d} p}{0} = -\frac{\mathrm{d} q}{0}.$$

Thus, from the last two fractions we can conclude that:

$$ p = A, \quad q = B, \quad A,B \in \mathbb{R}. $$

If you separately integrate both equations$^1$, since $p = \partial_x u$ and $q = \partial_y u$, you will get to:

\begin{align} u & = A x + \mu(y) \\ u & = B y + \eta(x) \end{align}

where $\mu$ and $\eta$ are some functions of their arguments. Looking at this set of conditions we can conclude that necesarily $\eta(x) = Ax$ and $\mu(y) = B$ and hence:

$$ \color{blue}{u(x,y) = A x+ By}$$

It's up to you now to show that any function of the form $\tilde{u} = u(x,y)+k$ is also a solution, where $k$ is a constant.

Cheers!

$^1$: Of course you can take advantage of the fact:

$$ \mathrm{d}u = p \, \mathrm{d}x + q \, \mathrm{d}y,$$

which is integrable since $p$ and $q$ are constants. Your result immediately follows.