Answer
$$z=4 r-5 s+2$$
Work Step by Step
Given $$F(r, s)=r^{2} s^{-1 / 2}+s^{-3}, \quad(2,1)$$
Since
\begin{align*}
F(r, s)&=r^{2} s^{-1 / 2}+s^{-3}\ \ \ \ & F(2,1)&=5 \\
F_r(r, s)&=2rs^{-1/2} \ \ \ \ & F_r(2,1)&= 4\\
F_s(r, s)&= \frac{-1}{2} r^2s^{-3/2}-3s^{-4} \ \ \ \ & F_s(2,1)&= -5 \\
\end{align*}
Then the tangent plane at $(2,1)$ is given by
\begin{align*}
z&= F(2,1)+ F_r(2,1)(r-2)+ F_s(2,1)(s-1)\\
&=5+4(r-2)-5(s-1)\\
&=4 r-5 s+2
\end{align*}
