Answer
$f_s=\frac{2t}{(s+t)^2}$
$f_t=\frac{-2s}{(s+t)^2}$
Work Step by Step
Take the first partial derivatives of the given function. When taking partial derivative with respect to s, treat t as a constant, and vice versa:
$f_s=\frac{(s+t)(1-0)-(s-t)(1+0)}{(s+t)^2}=\frac{s+t-s+t}{(s+t)^2}=\frac{2t}{(s+t)^2}$
$f_t=\frac{(s+t)(0-1)-(s-t)(0+1)}{(s+t)^2}=\frac{-s-t-s+t}{(s+t)^2}=\frac{-2s}{(s+t)^2}$
