I am given a function $u(x,y)$ such that $f(s,t) = u(e^{s+t} , e^{s-t})$
How can I prove that
$$ \frac{\partial^2 f}{\partial s \partial t} = 0 \iff x^2 \frac{\partial^2 u}{\partial x^2} - y^2 \frac{\partial^2 u}{\partial y^2} = y \frac{\partial u}{\partial y} - x \frac{\partial u}{\partial x} $$
I tried writing the derivative of $f$ according to $t$ and deriving the result according to $s$ but I did not get to the proof.
$$\frac{\partial f}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$$
The $s$ derivative of that gave me :
$$\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)\frac{\partial x}{\partial t}+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)\frac{\partial y}{\partial t}-\frac{\partial u}{\partial y}\frac{\partial^2y}{\partial s\partial t}+\frac{\partial u}{\partial x}\frac{\partial^2x}{\partial s\partial t}$$
I would really appreciate any help in this, thanks.
You mean $$f(s,t)=u(x,y), \text{ where }x=e^{},y=e^{s-t}$$ I'll use the notation $$u_1=\frac{\partial u}{\partial x},u_2=\frac{\partial u}{\partial y},u_{12}=\frac{\partial}{\partial y}(\frac{\partial u}{\partial x}) etc.$$Then $$\frac{\partial f}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$$ $$=u_1x+u_2{\partial y}y$$ Then $$\frac{\partial}{\partial t}(\frac{\partial f}{\partial s})=(u_{11}\frac{\partial x}{\partial t}+u_{12}\frac{\partial y}{\partial}t)x+u_1\frac{\partial x}{\partial t}$$ $$+(u_{21}\frac{\partial x}{\partial t}+u_{22}\frac{\partial y}{\partial t})y+u_2\frac{\partial y}{\partial t}$$ $$=(u_{11}x-u_{12}y)x+u_1x+(u_{21}x-u_{22}y)y-u_2y $$ $$=u_{11}x^2+u_1x-u_{22}y^2-u_2y$$ Thus $$\frac{\partial}{\partial t}(\frac{\partial f}{\partial s})=0$$ iff $$u_{11}x^2-u_{22}y^2=u_2y-u_1x.$$