Answer
The slope of the tangent line is $~\frac{4-3\sqrt{2}}{2}$
Work Step by Step
$r = 2+sin~3\theta$
$x = r~cos~\theta = (2+sin~3\theta)~(cos~\theta)$
$y = r~sin~\theta = (2+sin~3\theta)~(sin~\theta)$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3~cos~3\theta~sin~\theta+(2+sin~3\theta)~(cos~\theta)}{3~cos~3\theta~cos~\theta-(2+sin~3\theta)~(sin~\theta)}$
When $\theta = \frac{\pi}{4}$:
$\frac{dy}{dx} =\frac{3~cos~\frac{3\pi}{4}~sin~ \frac{\pi}{4}+(2+sin~\frac{3\pi}{4})~(cos~ \frac{\pi}{4})}{3~cos~\frac{3\pi}{4}~cos~ \frac{\pi}{4}-(2+sin~ \frac{3\pi}{4})~(sin~ \frac{\pi}{4})}$
$\frac{dy}{dx} =\frac{3~(-\frac{\sqrt{2}}{2})~(\frac{\sqrt{2}}{2})+(2+\frac{\sqrt{2}}{2})~(\frac{\sqrt{2}}{2})}{3~(-\frac{\sqrt{2}}{2})~(\frac{\sqrt{2}}{2})-(2+\frac{\sqrt{2}}{2})~(\frac{\sqrt{2}}{2})}$
$\frac{dy}{dx} =\frac{-\frac{3}{2}+\sqrt{2}+\frac{1}{2}}{-\frac{3}{2}-\sqrt{2}-\frac{1}{2}}$
$\frac{dy}{dx} =\frac{1-\sqrt{2}}{2+\sqrt{2}}$
$\frac{dy}{dx} =\frac{1-\sqrt{2}}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}$
$\frac{dy}{dx} =\frac{4-3\sqrt{2}}{2}$
The slope of the tangent line is $~\frac{4-3\sqrt{2}}{2}$
