<s>[INST] Solve $f'_x-xf'_y=y.$ 
Find the solution to $$\frac{\partial}{\partial x}f(x,y)
 -x\frac{\partial}{\partial y}f(x,y) = y$$
that satisfies $f(x,0)=x^2+\frac{x^3}{3},$ by using the transformation
$$\left\{   \begin{array}{rcr}
     u & = & ax^2+y \\
     v & = & x \\   \end{array} \right.$$
  for an appropriate value of $a$.

So I have that $f'_x-xf'_y=f'_x+vf'_y=y=u-ax^2=u-av^2.$ Furthermore:
\begin{array}{lcl}
u'_x & = & 2ax \\
u'_y & = & 1 \\
v'_x & = & 1 \\
v'_y & = & 0 \\
f'_x & = & f'_uu'_x+f'_vv'_x=2axf'_u+f'_v=2avf'_u+f'_v \\
f'_y & = & f'_uu'_y+f'_vv'_y=f'_u\cdot 1+f'_v\cdot 0=f'_u 
\end{array}
Plugging this into my PDE $f'_x-xf'_y=u-av^2$ I get
$$2avf'_u+f'_v-vf'_u = v(2af'_u-f'_u)+f'_v=u-av^2.$$
It's practical to choose $a=\frac{1}{2},$ then I get $$f'_v=u-\frac{v^2}{2} \ \implies f(u,v)=uv-\frac{v^3}{6} + g(u),$$
so
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right).$$
Using the condition $f(x,0)=x^2+\frac{x^3}{3}$ I get
$$f(x,0)=\frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right)= \Leftrightarrow g\left(\frac{x^2}{2}\right)=x^2.$$
Setting $t=\frac{x^2}{2}\Rightarrow x=\sqrt{2t}$ I get
$$g(t)=2t \implies g\left(\frac{x^2}{2}+y\right)= x^2+2y.$$
Finally:
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+x^2+2y.$$
But apparently this is wrong answer. Can anyone see where I made the error?
 [/INST] $$\begin{align}
f(x,y)= & \left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right)\\
f(x,0)= &\left(\frac{x^2}{2}+0\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+0\right) \\
=& \frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) = \frac{x^3}{3}+g\left(\frac{x^2}{2}\right) \neq \frac{x^2}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) \\
\end{align}$$
</s>