Given that u(s,t)=v(s^2-t^2,t^2-s^2) and v is differentiable prove that u verifies equation t*du/ds+s*du/dt=0

We'll note x = s^2 - t^2 and y = t^2 -
s^2

The constraint form enunciation u(s,t) =
v(s^2-t^2,t^2-s^2) is turning into u(s,t) = v(x,y).

We'll
apply the chain rule and we'll get:

du/ds = (dv/dx)*(dx/ds)
+ (dv/dy)*(dy/ds)

du/ds = (dv/dx)*(2s) + (dv/dy)*(-2s)
(1)

du/dt = (dv/dx)*(dx/dt) +
(dv/dy)*(dy/dt)

du/dt = (dv/dx)*(-2t) + (dv/dy)*(2t)
(2)

Now, we'll substitute (1) and (2) in the identity that
has to be demonstrated:

 t*(du/ds)+s*(du/dt) = [2st*(dv/dx)
- 2st(dv/dy)] + [-2st(dv/dx) + 2st(dv/dy)]

If
we'll remove the brackets, we'll cancel out the like terms and the
relation  t*(du/ds)+s*(du/dt) yields 0.