#### Answer

$$z=17+8(x-4)-2(y-1)$$

#### Work Step by Step

Given $$f(x, y)= x^2+y^{-2} \quad \text { at } \quad(4,1)$$
Since
\begin{aligned}
f_{x}(x, y) &=2x \\
f_{x}(4,1) &= 8
\end{aligned}
and
\begin{aligned}
f_{y}(x, y) &=-2y^{-3}\\
f_{y}(4,1) &=-2
\end{aligned}
Then the tangent plane given by
\begin{aligned}
z &=f(4,1)+f_{x}(4,1)(x-4)+f_{y}(4,1)(y-1) \\
&=17+8(x-4)-2(y-1)
\end{aligned}