Implicit differentiation to find dy/dx at a point

Differentiate both sides of $x^2y + y^3 = x + 5$ with respect to $x$. The term $x^2y$ requires the product rule: $\frac{d}{dx}[x^2y] = 2xy + x^2\frac{dy}{dx}$. The term $y^3$ requires the chain rule: $\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx}$. The right side gives $\frac{d}{dx}[x+5] = 1$. So we get $2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 1.$

Collect all terms containing $\frac{dy}{dx}$ on one side: $(x^2 + 3y^2)\frac{dy}{dx} = 1 - 2xy,$ so $\frac{dy}{dx} = \frac{1 - 2xy}{x^2 + 3y^2}.$ Substituting $x=1, y=1$ gives $\frac{dy}{dx} = \frac{1-2}{1+3} = \frac{-1}{4}$.