Solution attempt $xuu_x+yuu_y=u^2-1$ Solve $$
\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}
$$
I have got using Lagrange method:
$$F\left(\frac{x}{y}\right)=\frac{x^2}{u^2-1}$$
Applying $u(x,x^2)=x^3$:
$$u^2=\frac{y^6-x^6}{x^2y^2}+1$$
But plug in it to the PDE show that there is a mistake

Calling $v = u^2$ we have
$$
\frac 12 xv_x +\frac 12 y v_y = v-1
$$
with solution
$$
v = x^2\phi\left(\frac yx\right)+1
$$
now 
$$
v(x,x^2) = x^6\Rightarrow \phi\left(\frac yx\right) = \frac{y^6-x^6}{x^4y^2}
$$
and 
$$
v(x,y) = x^2\left(\frac{y^6-x^6}{x^4y^2}\right)+1
$$
and finally
$$
u(x,y) = \sqrt{\frac{y^6-x^6}{x^2y^2}+1}
$$
