Answer
$$\frac{d y}{d x}=-\frac{y}{x}$$
Work Step by Step
Given $$\sec x y+\tan x y+5=0$$
by letting $$F(x,y)=\sec x y+\tan x y+5=0=0$$
So, we have
$$F_x(x,y)=\frac{\partial F(x,y)}{\partial x}=y \sec x y \tan x y+y \sec ^{2} x y$$
$$F_y(x,y)=\frac{\partial F(x,y)}{\partial y}=x \sec x y \tan x y+x \sec ^{2} x y$$
Also, we get
\begin{aligned} \frac{d y}{d x}=&-\frac{F_{x}(x, y)}{F_{y}(x, y)} \\
&=-\frac{y \sec x y \tan x y+y \sec ^{2} x y}{x \sec x y \tan x y+x \sec ^{2} x y} \\ &=\frac{-y\left(\sec x y \tan x y+\sec ^{2} x y\right)}{x\left(\sec x y \tan x y+\sec ^{2} x y\right)}\\
&=-\frac{y}{x} \end{aligned}
