Original problem: find points where the tangent line is vertical or horizontal for the curve $ r^2 = \sin(2\theta)$. My work so far: $$ r^2 = \sin(2\theta) \\ 2r\frac{dr}{d\theta} = 2\cos(2\theta) \\ \frac{dr}{d\theta} = \frac{\cos(2\theta)}{r}$$
By the formula for the derivative of $y$ with respect to $x$: $$ \frac{dy}{dx} = \frac{r\cos(\theta) + \frac{dr}{d\theta} \sin(\theta)}{-r\sin(\theta) +\frac{dr}{d\theta} \cos(\theta)}$$ $$ = \frac{ r\cos(\theta) + \frac 1 r \cos(2\theta)\sin(\theta)} {-r\sin(\theta) + \frac 1 r \cos(2\theta)\cos(\theta)}$$
$$ \frac {dy}{dx} = \frac {\dfrac{r^2 \cos(\theta) + \cos(2\theta)\sin(\theta)}{r}} {\dfrac{-r^2 \sin(\theta) + \cos(2\theta)\cos(\theta)} {r}}$$
Can I cancel the $r$ without losing any points with horizontal or vertical tangents? How do I know if this is an acceptable operation (meaning that the operation does not alter the equation)? Any help is appreciated. Thank you