Does dy/dx change depending on the setup of the equation?

Question

So I noticed that when I found $dy/dx$ of $$x\sin(y)=1$$ I got $$dy/dx=-\tan(y)/x$$ where as when I try to find $dy/dx$ of $$\sin(y)=1/x$$ I get $$dy/dx=-\frac1{x^2\cos(y)}$$Shouldn't these two equations yield the same result since they are the same equation but algebraically manipulated before differentiating. When I drew the slope fields of the two graphs using a computer they yielded two different graphs. Why is this the case?

Answer 1

$$ 0 = \frac{d}{dx} (x \sin(y)) = \sin(y) + x \cos(y) \, y' \Rightarrow\\ y' = -\tan(y) / x $$ and $$ \sin(y) = 1/x \Rightarrow \\ \cos(y) \, y' = - 1/x^2 \Rightarrow \\ y' = - 1/x^2 \cdot 1/\cos(y) $$ You wrote $y' = -1/x^2 \cdot \cos(y)$ in your original post, before A.Γ.'s edit.

And indeed $$ y' = -1/x^2 \cdot 1/\cos(y) = - \sin(y)/x \cdot 1/\cos(y) = - \tan(y)/x $$