<s>[INST] 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
 [/INST] For the given PDE, Lagrange auxiliary equation $$\dfrac{dx}{xu}=\dfrac{dy}{yu}=\dfrac{du}{u^2-1}\tag1$$
From the first two ratio, $$\dfrac{dx}{xu}=\dfrac{dy}{yu}\implies \dfrac{dx}{x}=\dfrac{dy}{y}$$
Integrating we have $$\log x~=~\log y~+~\log c_1\implies\dfrac xy=c_1$$where $~c_1~$is a constant.
Again from the first and the last ratio, $$\dfrac{dx}{xu}=\dfrac{du}{u^2-1}\implies \dfrac{dx}{x}=\dfrac{u~du}{u^2-1}$$
Integrating we have$$2\log x~=~\log(u^2-1)~+~\log c_2\implies \dfrac{x^2}{u^2-1}~=~c_2$$where $~c_2~$is a constant.
Hence the general solution is $$F\left(\dfrac xy\right)=\dfrac{x^2}{u^2-1}\tag2$$where $~F~$is an arbitrary function.
Now given that $~u(x,x^2)=x^3~$ i.e., when $~y=x^2~$, then $~u=x^3~$. So from $(2)$, we have$$F\left(\dfrac 1x\right)=\dfrac{x^2}{x^6-1}$$
Hence $$F\left(\dfrac xy\right)=\dfrac{y^2/x^2}{y^6/x^6-1}=\dfrac{x^4y^2}{y^6-x^6}$$
From equation $(2)$ we have $$\dfrac{x^4y^2}{y^6-x^6}=\dfrac{x^2}{u^2-1}$$
$$\implies u^2-1=\dfrac{y^6-x^6}{x^2y^2}$$
$$\implies u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1$$This is the solution of the given PDE.

Cross-check : If possible let $~ u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1~\tag3$  is the solution of the given PDE $$\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}~.$$
Now putting $~y=x^2~$ in the solution we have 
$$~ u^2(x,x^2)=\dfrac{x^{12}-x^6}{x^2x^4}~+~1~=~x^{6}-1~+~1~=x^6\implies u(x,x^2)=x^3$$Hence the given condition is satisfied. 
Now we check whether the value satisfy the PDE or not ?
Differentiating the equation $(3)$ partially with respect to $~x~$, we have 
$$2uu_x=\dfrac{-6x^5}{x^2y^2}~-~2~\dfrac{y^6-x^6}{x^3y^2}=-2\dfrac{2x^6+y^6}{x^3y^2}$$ Again differentiating the equation $(3)$ partially with respect to $~y~$, we have 
$$2uu_y=\dfrac{6y^5}{x^2y^2}~-~2~\dfrac{y^6-x^6}{x^2y^3}=2\dfrac{x^6+2y^6}{x^2y^3}$$ 
Now $$xuu_x+yuu_y=-~\dfrac{2x^6+y^6}{x^2y^2}~+~\dfrac{x^6+2y^6}{x^2y^2}=\dfrac{-x^6+y^6}{x^2y^2}=u^2-1$$
Hence it is clear that 

$$~ u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1$$  is the solution of the given PDE $$\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}~.$$

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