Implicit differentiation to find dy/dx

Differentiate each term with respect to $x$. For $xy^2$, use the product rule: $\frac{d}{dx}[xy^2] = y^2 + x\cdot 2y\frac{dy}{dx}$. The term $-y$ differentiates to $-\frac{dy}{dx}$, and the right side $x^3$ differentiates to $3x^2$. This gives $y^2 + 2xy\frac{dy}{dx} - \frac{dy}{dx} = 3x^2.$

Collect the terms containing $\frac{dy}{dx}$ on the left side: $\frac{dy}{dx}(2xy - 1) = 3x^2 - y^2.$ Dividing both sides by $(2xy-1)$ isolates the derivative: $\frac{dy}{dx} = \frac{3x^2 - y^2}{2xy - 1}.$

Substitute $x=1$ and $y=2$ into $\frac{dy}{dx} = \frac{3x^2-y^2}{2xy-1}$: the numerator is $3(1)^2 - (2)^2 = 3-4=-1$, and the denominator is $2(1)(2)-1 = 3$. So the slope is $m = -\frac{1}{3}$.

Using point-slope form with the point $(1,2)$: $y - 2 = -\frac{1}{3}(x-1).$ Simplifying gives $y = -\frac{1}{3}x + \frac{7}{3}.$