SOLUTION Let \(x = s^2 -
t^2\) and \(y = t^2 - s^2\).
Then \(g(s, t) = f(x, y)\) and the
Chain Rule gives \[ \frac{\partial
g}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial
x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial
y}{\partial s} = \frac{\partial f}{\partial x}(2s) + \frac{\partial
f}{\partial y}(-2s) \] \[
\frac{\partial g}{\partial t} = \frac{\partial f}{\partial x}
\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}
\frac{\partial y}{\partial t} = \frac{\partial f}{\partial x}(-2t) +
\frac{\partial f}{\partial y}(2t) \] Therefore \[ t \frac{\partial g}{\partial s} + s
\frac{\partial g}{\partial t} = t \left( 2s \frac{\partial f}{\partial
x} - 2s \frac{\partial f}{\partial y} \right) + s \left( -2t
\frac{\partial f}{\partial x} + 2t \frac{\partial f}{\partial y} \right)
= 0 \]