Under the transformation $ \xi=u_x(x,y) , \ \eta=u_y(x,y) \ and \ \ \phi=xu_x+yu_y-u \ $ Under the transformation $ \xi=u_x(x,y) , \ \eta=u_y(x,y) \ and \ \ \phi=xu_x+yu_y-u \ $ , Show that the Quasi-linear equation $ \ a(u_x,u_y) u_{xx} +2b(u_x,u_y) u_{xy} +c(u_x,u_y) u_{yy}=0 \ $ transforms to the linear equation, 
$$ a(\xi, \eta) \phi_{\eta \eta} -2b (\xi, \eta) \phi_{\xi \eta} +c(\xi, \eta) \phi_{\xi \xi}= 0 $$
My Attempt:
Given, 
$$\xi=u_x(x,y) , \qquad \eta=u_y(x,y)$$
Then, 
$$ \phi=x \xi+y \eta-u \ $$ 
Now, 
$$u_x=\xi \  \implies \ \  u_{xx}=u_{\xi} \xi_{x}+u_{\eta} \eta_{x}$$
But now I can't proceed further.
Help me out of this 
 A: I just learned about this from the following book, page 121. There is also a small section on this in page 210 of Evans, Partial Differential Equations, Second Edition. I reproduce the calculation for your equation below. There is apparently a link to Legendre transforms but I am not well versed in the topic so I will leave it at that.
By differentiating wrt $\xi$ the equation $\xi = u_x$ with the chain rule, $$\xi = u_x ⟹ 1=u_{xx}x_\xi+u_{xy}y_\xi$$
We also note that $$\phi_\xi=x+\xi x_\xi +y_\xi\eta- \xi x_\xi - \eta y_\xi = x $$
and hence, $x_\xi = \phi_{\xi\xi}$. Similarly, $y_\xi = \phi_{\eta\xi} $. This means that we can write
$$ 1 = u_{xx}\phi_{\xi\xi} + u_{xy}\phi_{\eta\xi}$$
Differentiating wrt $\eta$, and also repeating for the other equation $\eta = u_y$ we obtain 4 equations which can be written as the matrix equation
$$ \begin{pmatrix} u_{xx} & u_{yx} \\ u_{xy} & u_{yy} \end{pmatrix} \begin{pmatrix} \phi_{\xi\xi} & \phi_{\eta\xi} \\ \phi_{\xi\eta} & \phi_{\eta\eta} \end{pmatrix} =  \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
From here it is easy to invert the right matrix on the left-hand-side (which by the way, needs to be invertible for the change of variables to make sense) to obtain 
$$ \begin{pmatrix} u_{xx} & u_{xy} \\ u_{xy} & u_{yy} \end{pmatrix}  = \frac1{\phi_{\xi\xi}\phi_{\eta\eta} - \phi_{\eta\xi}\phi_{\xi\eta}}  \begin{pmatrix} \phi_{\eta\eta} & -\phi_{\xi\eta} \\ -\phi_{\eta\xi} & \phi_{\xi\xi} \end{pmatrix}$$
Substituting this into the equation gives the result.
