To find $\frac{dy}{dx}$ using implicit differentiation for the equation in Problem 21, $xy(2x + 3y) = 2$, follow these steps:

1. Differentiate both sides of the equation with respect to $x$:

   The left side is $xy(2x + 3y)$. We need to use the product rule and the chain rule.

   $\frac{d}{dx}[xy(2x + 3y)] = \frac{d}{dx}(xy) \cdot (2x + 3y) + xy \cdot \frac{d}{dx}(2x + 3y)$

   Differentiate $xy$:

   Using the product rule: $u = x$ and $v = y$, so $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.

   Differentiate $2x + 3y$:

   $\frac{d}{dx}(2x + 3y) = 2 + 3\frac{dy}{dx}$.

   Substitute back into the equation:

   $(x \frac{dy}{dx} + y)(2x + 3y) + xy(2 + 3\frac{dy}{dx}) = 0$

2. Expand and combine terms:

   $(2x^2 + 3xy + 2x(y \frac{dy}{dx}) + 3y^2 \frac{dy}{dx}) + 2xy + 3xy \frac{dy}{dx} = 0$

3. Combine like terms:

   $2x^2 + 5xy + (2xy + 3y^2 + 3xy)\frac{dy}{dx} = 0$

4. Isolate $\frac{dy}{dx}$:

   $(2xy + 3y^2 + 3xy)\frac{dy}{dx} = -2x^2 - 5xy$

   $\frac{dy}{dx} = \frac{-2x^2 - 5xy}{2xy + 3y^2 + 3xy}$

   Simplify if necessary.

This gives you the expression for $\frac{dy}{dx}$.