Find the points on the given curve where the tangent line is horizontal or vertical....

To find the points where the tangents are horizontal/vertical, we need to find dy/dx. Using the chain rule fore differentiation, we can write
${\displaystyle \frac{dy}{dx}=\frac{dy}{d\theta }\cdot \frac{d\theta }{dx}=\frac{\frac{dy}{d}\theta }{\frac{dx}{d}\theta }}$
Substitute ${\displaystyle y=r\sin \theta =e^{\theta }\sin \theta }$ and $x=r\cos \theta =e^{\theta }$, to get
${\displaystyle \frac{dy}{dx}=\frac{d\frac{e^{\theta }\sin \theta }{d}\theta }{d\frac{e^{\theta }\cos \theta }{d}\theta }}$
Use product rule for differentiation
${\displaystyle \frac{dy}{dx}=\frac{d{\left(e^{\theta }\right)}^{\prime }\sin \theta +e^{\theta }{\left(\sin \theta \right)}^{\prime }}{{\left(e^{\theta }\right)}^{\prime }\cos \theta +e^{\theta }{\left(\cos \theta \right)}^{\prime }}}$
${\displaystyle \frac{dy}{dx}=\frac{e^{\theta }\sin \theta +e^{\theta }\cos \theta }{e^{\theta }\cos \theta -e^{\theta }\sin \theta }}$
Divide both the numerator and the denominator by ${\displaystyle e^{\theta }}$
${\displaystyle \frac{dy}{dx}=\frac{\sin \theta +\cos \theta }{\cos \theta -\sin \theta }}$
The tangents will be horizontal when the numerator is 0
${\displaystyle \sin \theta +\cos \theta =0}$
Subtract ${\displaystyle \cos \theta }$ from both sides
${\displaystyle \sin \theta =-\cos \theta }$
Divide both sides by ${\displaystyle \cos \theta }$
${\displaystyle \frac{\sin \theta }{\cos \theta }=-1}$
In the right-hand side, we will rewrite -1 as ${\displaystyle \tan (-\frac{\pi }{4})}$
${\displaystyle tam\theta =\tan (-\frac{\pi }{4})}$
Remember that: the general solution of the equation ${\displaystyle \tan x=\tan \alpha }$ is
${\displaystyle x=\alpha +n\pi }$
Therefore, the tangents are horizontal when
${\displaystyle \theta =-\frac{\pi }{4}+n\pi }$
The tangents will be vertical when the denomiantor is 0
${\displaystyle \sin \theta =\cos \theta }$
Divides both sides by ${\displaystyle \cos \theta }$
${\displaystyle \frac{\sin \theta }{\cos \theta }=1}$
In the right-hand side, we will rewrite 1 as ${\displaystyle \tan \left(\frac{\pi }{4}\right)}$
${\displaystyle \tan \theta =\tan \left(\frac{\pi }{4}\right)}$
Therefore, the tangents are vertical when