partial derivative of $f(x,y)$ who satisfies $f_{xx}-f_{yy}=0$ Suppose that $z=f(x,y)$ and its second-order partial derivative is continuous. It also satisfies $\displaystyle\frac{\partial^{2}f}{\partial x^{2}}-\frac{\partial^{2}f}{\partial y^{2}}=0$,$f(x,2x)=x$ and $f_{x}^{\prime}(x,2x)=x^2$. So how can we obtain
$$\displaystyle f_{y}^{\prime}(x,2x),f_{xx}^{\prime\prime}(x,2x),f_{yy}^{\prime\prime}(x,2x),f_{xy}^{\prime\prime}(x,2x)?$$
 A: Can you show that the solution of the PDE $f_{xx} - f_{yy} = 0$ is $f(x,y) = F(x+y) + G(x-y)$, where $F$ and $G$ are arbitrary functions of their arguments?
Edit (how to get to the solution):
Consider the equation $f_{xx} - f_{yy} = 0$, which can be written in operator form as: $$(D_{x} - D_y )(D_{x}+D_y) \, f = 0, \quad D_x = \frac{\partial}{\partial x}, \quad D_y = \frac{\partial}{\partial y}. $$
Define $(D_x+D_y) \, f \equiv u $ to get: 
$$u_x - u_y = 0,$$ which is a linear 1st order PDE and solvable by the method of characteristics, yielding:
$$ \frac{\mathrm{d} x}{1} = \frac{\mathrm{d}y}{-1} = \frac{\mathrm{d}u}{0} \implies  \; x+ y= c_1, \quad u =  c_2. $$ Now you can put $c_2 = g(c_1)$ to get $u = g(x+y)$, where $g$ is an arbitrary function of $x+y$. With this information, we finally get:
$$f_x + f_y = u = g(x+y),$$ which can be solved in a similar fashion, i.e.:
$$ \frac{\mathrm{d} x}{1} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}u}{0} \implies  \; x - y= d_1  $$ and, taking for example $x = d_1 +y$, we'd get $ \mathrm{d}u =g(d_1+2y)  \; \mathrm{d}y $, which integrates to:
$$ u = G(d_1+2y) + d_2,$$ where $G$ is a new arbitrary function of its arguments. Put now $d_2$ as a function of $d_1$ to finally have:

$$ \color{blue}{u(x,y) = G(x+y) + F(x-y)}$$

The conditions for $u$ require the following to hold:
\begin{align}
 f(x,2x)   & = x = G(3x) + F(-x), \\ 
 f_x(x,2x) & = x^2 = G'(3x) + F'(-x)
\end{align}
What would happen if you differentiate the first equation wrt $x$ and sum it up to the second? Can you take it from here?
Hope this helps.
Cheers!

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