Answer
$$y'=\frac{y}{x}$$
Work Step by Step
$$\ln(x/y)=1$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$(\ln(x/y))'=0$$
Recall that $(\ln x)'=1/x$: $$\frac{1}{\frac{x}{y}}\times\Big(\frac{x}{y}\Big)'=0$$
$$\frac{y}{x}\times\frac{x'y-xy'}{y^2}=0$$
$$\frac{y}{x}\times\frac{y-xy'}{y^2}=0$$
$$\frac{y-xy'}{xy}=0$$
Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation:
$$\frac{y}{xy}-\frac{xy'}{xy}=0$$
$$\frac{1}{x}-\frac{y'}{y}=0$$
$$\frac{y'}{y}=\frac{1}{x}$$
Finally, calculate for $y'$: $$y'=\frac{1}{x}\times y=\frac{y}{x}$$
