1. The first problem is to find the partial derivatives $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$ for the function $f(x,y) = \frac{y}{x^2 + y^2}$ where $x = r \cos \theta$ and $y = r \sin \theta$, and then calculate their product. 2. We start by substituting $x$ and $y$ in terms of $r$ and $\theta$: $$f(r,\theta) = \frac{r \sin \theta}{(r \cos \theta)^2 + (r \sin \theta)^2} = \frac{r \sin \theta}{r^2 (\cos^2 \theta + \sin^2 \theta)} = \frac{r \sin \theta}{r^2} = \frac{\sin \theta}{r}$$ 3. Now, compute the partial derivatives: - Partial derivative with respect to $r$: $$\frac{\partial f}{\partial r} = \frac{\partial}{\partial r} \left( \frac{\sin \theta}{r} \right) = -\frac{\sin \theta}{r^2}$$ - Partial derivative with respect to $\theta$: $$\frac{\partial f}{\partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{\sin \theta}{r} \right) = \frac{\cos \theta}{r}$$ 4. Finally, calculate the product: $$\frac{\partial f}{\partial r} \cdot \frac{\partial f}{\partial \theta} = \left(-\frac{\sin \theta}{r^2}\right) \cdot \left(\frac{\cos \theta}{r}\right) = -\frac{\sin \theta \cos \theta}{r^3}$$ **Answer:** $$\boxed{-\frac{\sin \theta \cos \theta}{r^3}}$$