Answer
$$y = -\frac{12}{5} x+\frac{32}{5}$$
Work Step by Step
Given $$2 x^{1 / 2}+4 y^{-1 / 2}=x y,\ \ \ (1,4)$$
Differentiate both sides with respect to $x$
\begin{align*}
\frac{d}{dx}(2 x^{1 / 2}+4 y^{-1 / 2} )&=\frac{d}{dx}(x y )\\
x^{-\frac{1}{2}}-2 y^{-\frac{3}{2}} y^{\prime}&=x y^{\prime}+y\\
\left(-2 y^{-\frac{3}{2}}-x\right) y^{\prime}&=y-x^{-\frac{1}{2}}\\
y^{\prime}&=\frac{y-x^{-\frac{1}{2}}}{-2 y^{-\frac{1}{2}}-x}
\end{align*}
Then at $(1,4)$
\begin{align*}
m&= \frac{4-(1)^{-\frac{1}{2}}}{-2(4)^{-\frac{3}{2}}-1}\\
&= \frac{-12}{5}
\end{align*}
Then the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-4}{x-1}&= \frac{-12}{5}\\
y-4 &=-\frac{12}{5} \cdot(x-1) \\
y &=-\frac{12}{5} x+\frac{12}{5}+4\\
y&= -\frac{12}{5} x+\frac{32}{5}
\end{align*}
