\( \begin{array}{r}\text { Find } d y / d x \text { by implicit differentiation. } \\ \tan ^{-1}\left(3 x^{2} y\right)=x+2 x y^{2} \\ y^{\prime}=\square\end{array} \)

To find \( \frac{dy}{dx} \) using implicit differentiation, start by differentiating both sides of the equation \( \tan^{-1}(3x^2y) = x + 2xy^2 \): 1. Differentiate the left side using the chain rule: \[ \frac{d}{dx}\left(\tan^{-1}(3x^2y)\right) = \frac{1}{1 + (3x^2y)^2} \cdot \frac{d}{dx}(3x^2y) = \frac{1}{1 + 9x^4y^2}\left(3(2xy + x^2\frac{dy}{dx})\right) = \frac{6xy + 3x^2y'}{1 + 9x^4y^2} \] 2. Differentiate the right side: \[ \frac{d}{dx}(x + 2xy^2) = 1 + 2y^2 + 2x(2y\frac{dy}{dx}) = 1 + 2y^2 + 4xy\frac{dy}{dx} \] Setting both derivatives equal gives: \[ \frac{6xy + 3x^2y'}{1 + 9x^4y^2} = 1 + 2y^2 + 4xy\frac{dy}{dx} \] Now, clear the fraction by multiplying through by the denominator: \[ 6xy + 3x^2y' = (1 + 2y^2 + 4xy\frac{dy}{dx})(1 + 9x^4y^2) \] This will yield an equation that can be rearranged to solve for \( y' \). After some algebraic manipulation, isolate \( y' \) terms on one side and factor out \( y' \): \[ y'(3x^2 - 4xy(1 + 9x^4y^2)) = \text{other terms} \] Thus, we can express \( y' \) as: \[ y' = \frac{\text{other terms}}{3x^2 - 4xy(1 + 9x^4y^2)} \] So, the value of \( \frac{dy}{dx} \) will depend on the simplified other terms from the right-hand side equation substitution. Pitching the right combination can lead you to the final value to fill in the square!