Let us suppose we want to find the curve which is perpendicular to the curve $y=3x$. From the concept of straight lines,we know that the slope of that curve will be $-\frac{1}{3}$. Whose equation should be $y=-\frac{1}{3}x+c$.This can also be proved using differential equations. The derivative of 2nd curve will be $-\frac{1}{3}$. Hence, $\frac{dy}{dx}=-\frac {1}{3}$. By separation of variables we get $dy=-\frac{1}{3}dx$ and upon integration we get $y=-\frac{1}{3}x+c$. But let's see it from a different perspective.
The differential equation of first curve is $\frac{dy}{dx}=3=\frac{y}{x}$. Now,to get the orthogonal curve,we will have to replace $\frac{dy}{dx}$ by $-\frac{1}{\frac{dy}{dx}}$. And so the differential equation of the 2nd curve is $\frac{dy}{dx}=-\frac{x}{y}$. Again by variable separation we get,$ydy=-xdx$ which upon integration we get $\frac{y^2}{2}=-\frac{x^2}{2}+c$ which is $x^2+y^2=k$. Here we are getting straight line as the answer and again circle as the answer which is a contradiction. Although it is obvious and common sense that the straight line method is correct but I don't understand what's wrong with the circle method. It will be really helpful if I am pointed out where I made a fallacy.
The differential equation that defines your first curve cannot be uniquely defined as $\frac{y}{x}$; that is a property of all lines. The slope is simply $3$ everywhere: $\frac{dy}{dx}=3$. Thus, when you apply your method, you get all possible perpendicular lines. Since the first equation defines a radius of a circle centered at the origin, your perpendicular lines are tangents to the circle.