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