1. **State the problem:** We need to find the equation of a curve such that at every point $(x,y)$ on the curve, the slope of the tangent line is equal to $-\frac{y}{x+y}$. 2. **Write the slope condition as a differential equation:** The slope of the tangent line is the derivative $\frac{dy}{dx}$. So, we have $$\frac{dy}{dx} = -\frac{y}{x+y}.$$ 3. **Rewrite the equation:** Multiply both sides by $x+y$ to get $$(x+y)\frac{dy}{dx} = -y.$$ 4. **Separate variables or rearrange:** Rewrite as $$(x+y)dy = -y dx.$$ 5. **Express in terms of $x$ and $y$ differentials:** $$(x+y)dy + y dx = 0.$$ 6. **Check if the differential form is exact:** Let $$M = y, \quad N = x + y.$$ Calculate partial derivatives: $$\frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = 1.$$ Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact. 7. **Find potential function $F(x,y)$ such that:** $$\frac{\partial F}{\partial x} = y, \quad \frac{\partial F}{\partial y} = x + y.$$ 8. **Integrate $M = y$ with respect to $x$:** $$F(x,y) = xy + h(y),$$ where $h(y)$ is an arbitrary function of $y$. 9. **Differentiate $F$ with respect to $y$ and equate to $N$:** $$\frac{\partial F}{\partial y} = x + h'(y) = x + y.$$ So, $$h'(y) = y \implies h(y) = \frac{y^2}{2} + C.$$ 10. **Write the implicit solution:** $$F(x,y) = xy + \frac{y^2}{2} = C,$$ where $C$ is a constant. **Final answer:** The equation of the curve is $$xy + \frac{y^2}{2} = C.$$