$$ yu_x - 3x^2yu_y = 3x^2u $$
I then get my $\frac{dy}{dx}$ = $-3x^2$ which leads to $ \phi = y+x^3$ and $\xi = x$
I then find $u_x$ and $u_y$ by using the chain rule.
$$ u_x = \frac{du}{d\phi} \frac{d\phi}{dx} + \frac{du}{d\xi} \frac{d\xi}{dx}$$ $$ u_y = \frac{du}{d\phi} \frac{d\phi}{dy} + \frac{du}{d\xi} \frac{d\xi}{dy}$$ $$ u_x = 3x^2\frac{du}{d\phi} + \frac{du}{d\xi} $$ $$ u_y = \frac{du}{d\phi} $$
Substituting these into the equation above
$$ 3x^2y\frac{du}{d\phi} + y\frac{du}{d\xi} - 3x^2y\frac{du}{d\phi} = 3x^2u $$ $$ y\frac{du}{d\xi} = 3x^2u $$
it is at this pint where i get stuck and i am not really sure how to move forward with the problem. My first thought is to change y to $\phi - x^3$ and then say since $x=\xi$ therefore $y=\phi - \xi^3$ and therefore $3x^2u = 3\xi^2u$ but i couldn't get much further.
Apologies if similar questions have been asked, i read a few but i didn't really understand the method they were using (not saying they were wrong, im just not good enough to understand it).
Characteristics can be found from the following system of equations
$$ \frac{dx}{y} = \frac{dy}{-3x^2y} = \frac{du}{3x^2u} $$
Directly solving it one may obtain
$$ \begin{aligned} \phi_1(x, y, u) &= y+x^3 = c_1\\ \phi_2(x, y, u) &= u y = c_2 \end{aligned} $$
General solution of PDE is written in terms of $\phi_1$ and $\phi_2$
$$ \begin{aligned} F(\phi_1, \phi_2) = 0\\ F(y+x^3, u y) = 0\\ u y &= f(y+x^3)\\ u &= \frac{1}{y}f(y+x^3) \end{aligned} $$
where $F$ and $f$ are arbitrary functions.