1. **State the problem:** We want to evaluate the double integral $$\iint_R \frac{-y}{x^2 y^2 + 1} \, dA$$ where the region $$R$$ is the unit square defined by $$0 \leq x \leq 1$$ and $$0 \leq y \leq 1$$. 2. **Set up the integral:** Since $$R$$ is a rectangle, we can write the double integral as an iterated integral: $$\int_0^1 \int_0^1 \frac{-y}{x^2 y^2 + 1} \, dy \, dx$$. 3. **Integrate with respect to $$y$$ first:** Fix $$x$$ and consider $$I(x) = \int_0^1 \frac{-y}{x^2 y^2 + 1} \, dy$$. 4. **Use substitution:** Let $$u = x^2 y^2 + 1$$, then $$du = 2 x^2 y \, dy \implies y \, dy = \frac{du}{2 x^2}$$. Rewrite the integral: $$I(x) = - \int_0^1 \frac{y}{x^2 y^2 + 1} \, dy = - \int_{u=1}^{u=x^2 + 1} \frac{1}{u} \cdot \frac{du}{2 x^2} = - \frac{1}{2 x^2} \int_1^{x^2 + 1} \frac{1}{u} \, du$$. 5. **Integrate:** $$I(x) = - \frac{1}{2 x^2} [\ln(u)]_1^{x^2 + 1} = - \frac{1}{2 x^2} \ln(x^2 + 1)$$. 6. **Now integrate with respect to $$x$$:** $$\int_0^1 I(x) \, dx = - \frac{1}{2} \int_0^1 \frac{\ln(x^2 + 1)}{x^2} \, dx$$. 7. **Evaluate the integral $$J = \int_0^1 \frac{\ln(x^2 + 1)}{x^2} \, dx$$:** This integral is improper at $$x=0$$ because of $$\frac{1}{x^2}$$, but the logarithm behaves well. We use integration by parts: Let $$u = \ln(x^2 + 1)$$, $$dv = \frac{1}{x^2} dx$$. Then $$du = \frac{2x}{x^2 + 1} dx$$, and $$v = -\frac{1}{x}$$. Integration by parts gives: $$J = uv \bigg|_0^1 - \int_0^1 v \, du = \left[-\frac{\ln(x^2 + 1)}{x}\right]_0^1 + \int_0^1 \frac{1}{x} \cdot \frac{2x}{x^2 + 1} dx = -\ln(2) + \lim_{x \to 0} \frac{\ln(x^2 + 1)}{x} + 2 \int_0^1 \frac{1}{x^2 + 1} dx$$. The limit $$\lim_{x \to 0} \frac{\ln(x^2 + 1)}{x} = 0$$ by L'Hôpital's rule. The remaining integral is: $$2 \int_0^1 \frac{1}{x^2 + 1} dx = 2 [\arctan(x)]_0^1 = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$$. So, $$J = -\ln(2) + \frac{\pi}{2}$$. 8. **Substitute back:** $$\int_0^1 I(x) dx = - \frac{1}{2} J = - \frac{1}{2} \left(-\ln(2) + \frac{\pi}{2}\right) = \frac{1}{2} \ln(2) - \frac{\pi}{4}$$. **Final answer:** $$\boxed{\frac{1}{2} \ln(2) - \frac{\pi}{4}}$$