Then, it is a simple exercise to show that:$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{f'(\theta)sin(\theta)+f(\theta)cos(\theta)}{f'(\theta)cos(\theta)-f(\theta)sin(\theta)}$$
Finally, since $f'(\theta)=\frac{d}{d\theta}e^{\theta}=e^{\theta}=f(\theta)$ this simplifies to:$$\frac{dy}{dx}=\frac{sin(\theta)+cos(\theta)}{cos(\theta)-sin(\theta)}$$
So, when is this final equation equal to $0$? When is it indeterminate? And what does this imply about the slope of the tangent line?
Your first steps seem right, but then it is much more natural to solve for $y$: $~~y = -x^2$. Of course, the points have to lie on the original curve, so substitute this back into the original equation to get
$$
2x^4 - 2x^3 - 4x^2 - 1 = 0 ~.
$$
So now you need the real roots of this quartic. This is slightly tricky though; it certainly has real roots, but I don't think it has any rational roots (assume it does, derive a contradiction), for example...
Best Answer
You are given $r$ as a function of $\theta$,i.e., $r=f(\theta)$, so you rewrite the usual identities: $$x = rcos(\theta)$$ $$y=rsin(\theta)$$
As: $$x=f(\theta)cos(\theta)$$ $$y=f(\theta)sin(\theta)$$
Then, it is a simple exercise to show that:$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{f'(\theta)sin(\theta)+f(\theta)cos(\theta)}{f'(\theta)cos(\theta)-f(\theta)sin(\theta)}$$
Finally, since $f'(\theta)=\frac{d}{d\theta}e^{\theta}=e^{\theta}=f(\theta)$ this simplifies to:$$\frac{dy}{dx}=\frac{sin(\theta)+cos(\theta)}{cos(\theta)-sin(\theta)}$$
So, when is this final equation equal to $0$? When is it indeterminate? And what does this imply about the slope of the tangent line?