Answer
$$\frac{{dr}}{{d\theta }} = -\frac{{r + {{\csc }^2}\theta }}{{\sin r + \theta }}$$
Work Step by Step
$$\eqalign{
& \cos r + \cot \theta = r\theta \cr
& {\text{differentiate both sides with respect to }}\theta \cr
& \frac{d}{{d\theta }}\left( {\cos r} \right) + \frac{d}{{d\theta }}\left( {\cot \theta } \right) = \frac{d}{{d\theta }}\left( {r\theta } \right) \cr
& \cr
& {\text{use the product rule}} \cr
& \frac{d}{{d\theta }}\left( {\cos r} \right) + \frac{d}{{d\theta }}\left( {\cot \theta } \right) = r\frac{d}{{d\theta }}\left( \theta \right) + \theta \frac{d}{{d\theta }}\left( r \right) \cr
& \cr
& {\text{find derivatives}} \cr
& - \sin r\frac{{dr}}{{d\theta }} - {\csc ^2}\theta = r + \theta \frac{{dr}}{{d\theta }} \cr
& {\text{Solve for }}\frac{{dr}}{{d\theta }} \cr
& -\sin r\frac{{dr}}{{d\theta }} - \theta \frac{{dr}}{{d\theta }} = r + {\csc ^2}\theta \cr
& \left( {-\sin r - \theta } \right)\frac{{dr}}{{d\theta }} = r + {\csc ^2}\theta \cr
& \frac{{dr}}{{d\theta }} = \frac{{r + {{\csc }^2}\theta }}{{-\sin r - \theta }} \cr} $$
