Finding the horizontal and vertical tangents of a parametric equation. 
Find the points at which the polar curve $r=2+2\sin{(\theta)}$ has a
  horizontal or vertical tangent line.

Translate the parametric equation to Cartesian coordinates:
$$
r^2=2r+2r\sin{(\theta)} \\
x^2+y^2=2\sqrt{x^2+y^2}+2y \\
x^2+y^2-2y+1=2\sqrt{x^2+y^2}+1 \\
x^2+(y-1)^2=2\sqrt{x^2+y^2}+1 \\
$$
Then differentiate with respect to $x$:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\
2x+2(y-1)y'=\frac{2(x+y)}{\sqrt{x^2+y^2}} \\
x+(y-1)y'=\frac{x+y}{\sqrt{x^2+y^2}}y' \\
y'=\frac{x}{\frac{x+y}{\sqrt{x^2+y^2}}-(y-1)} \\
y'=\frac{x}{\frac{x+y-\sqrt{x^2+y^2}(y-1)}{\sqrt{x^2+y^2}}} \\
y'=\frac{x\sqrt{x^2+y^2}}{x+y-\sqrt{x^2+y^2}(y-1)} \\
$$
So, by now it seems I'm not making any real progress in finding the vertical and horizontal tangents of this equation.
Am I going about this wrong?

Mistake in taking derivatives:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2+(y-1)^2\right]
     =\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{x^2+y^2}+1\right] \\
2x+2(y-1)y'=\frac{2x+2yy'}{\sqrt{x^2+y^2}}
$$
and you dropped the $y'$ on the RHS...
UPDATE
Another approach is to let $x=x(\theta)$ and $y = y(\theta)$ and compute
$$
\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}
$$
In our case, $r=2+2\sin \theta$ so $$x = r\cos \theta = \cos \theta (2+2\sin \theta)$$ and similarly $$y = r\sin \theta = \sin \theta (2+2\sin \theta)$$, so computing $y'(\theta)$ is straight-forward arithmetic.
Setting it to $0$ and $\pm \infty$ should give you the relevant angles...
