Answer
\begin{align*}
(a)&\frac{\partial h}{\partial s} =\frac{1}{t}-2t(s t-t r) \\
(b)&\frac{\partial h}{\partial t} =\frac{-s}{t^2}+\frac{1}{r}-2(s t-t r)(s-r) \\
(c)&\frac{\partial h}{\partial r} = \frac{-t}{r^2}+2t(s t-t r) \\
(d)&\frac{\partial h}{\partial r}\bigg|_{ ( 1,2,-1)} = \frac{-2}{(-1)^2}+2(2)(2+2) = 14\\
\end{align*}
Work Step by Step
Given $$h(s, t, r)=\frac{s}{t}+\frac{t}{r}-(s t-t r)^{2}$$
Since
\begin{align*}
(a)&\frac{\partial h}{\partial s} =\frac{1}{t}-2t(s t-t r) \\
(b)&\frac{\partial h}{\partial t} =\frac{-s}{t^2}+\frac{1}{r}-2(s t-t r)(s-r) \\
(c)&\frac{\partial h}{\partial r} = \frac{-t}{r^2}+2t(s t-t r) \\
(d)&\frac{\partial h}{\partial r}\bigg|_{ ( 1,2,-1)} = \frac{-2}{(-1)^2}+2(2)(2+2) = 14\\
\end{align*}
