Chain rule in multi-variable functions 
$u(x,y)\in c^2  $ in $  D=\{(0,\infty) \times(0,\infty)\}, v(s,t)=u(e^{s+t},e^{s-t})$. Prove that $$v_{st}=0 \iff x^2 u_{xx}+x u_x=y^2 u_{yy}+y u_y$$

So what I am trying to do it to find $v_{st}$.
I will define: $x(s,t)=e^{s+t},y(s,t)=e^{s-t}$.
$(x(s,t),y(s,t))\in D $ for all $  (s,t)\in \mathbb{R}^2$, so we can use the chain rule twice.
$$v_s=e^{s+t}u_x+e^{s-t}u_y=xu_x+yu_y$$
Now this is the part where it gets tricky to me.
${v_s}_t=(xu_x+yu_y)_t$
$(xu_x)_t=x_tu_x+x{u_x}_t$. 
Now I am stuck, as I am not sure how to handle ${u_x}_t$.
Apparently, I need to apply the chain rule on $u_x$ but I don't understand why.
Moreover, after applying the chain rule I am supposed to get ${u_x}_t={u_x}_xx_t+{u_x}_yy_t$. Even after applying the chain rule I can't understand how do I reach this. 
Can someone explain those two parts to me?
Thanks in advance.
 A: You've defined $u=u(x,y)$ and $x=x(s, t), y=y(s, t)$ and because $u$ is a function of $x$ and $y$, so is $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}(x,y).$
The chain rule is
$$\frac{\partial}{\partial t} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial x} \right) \frac{\partial x}{\partial t} +\frac{\partial}{\partial y} \left( \frac{\partial u}{\partial x} \right) \frac{\partial y}{\partial t},$$
which is really just what you wrote in an alternate notation:
$$u_{xt} = u_{xx} x_t + u_{xy} y_t.$$
UPDATE:
I think it is a matter of going step-by-step ...
As you've defined:  
$$x=e^{s+t}, \quad y=e^{s-t},$$ so that $$x_s = x, \quad x_t = x, \quad y_s = y, \quad y_t = -y.$$
Since $v=u(e^{s+t},e^{s-t}),$ $$v=u(x(s,t),y(s,t)).$$
Now $$v_s = u_x x_s + u_y y_s = x u_x + y u_y$$
$$v_{st} = x_t u_x + x u_{xt} + y_t u_y + y u_{yt}$$
$$v_{st} = xu_x -yu_y + x u_{xx} + y u_{yt}$$
$$v_{st} = xu_x = y u_y + x[u_{xx} x_t + u_{xy} y_t ] + y[u_{yx}x_t + u_{yy} y_t].$$
$$v_{st} = xu_x -yu_y  + x[u_{xx} x - u_{xy} y]+ y[u_{yx} x - u_{yy} y].$$
$$v_{st} = x^2 u_{xx} - y^2 u_{yy} + xu_x -y u_y, $$ because we assume that the derivatives are sufficiently nice (continuous) so $u_{xy} = u_{yx}$.
Since $v_{st}=0$ is equivalent to the left-hand side minus the right-hand side being equal to zero, we have shown that $v_{st}=0$ if and only if $x^2 u_{xx} + x u_x = y^2 u_{yy} + y u_y.$
