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

We start with the equation \[ \tan^{-1}(3x^2y) = x + 2xy^2. \] Differentiate both sides with respect to \( x \). 1. Differentiate the left-hand side using the chain rule. Recall that \[ \frac{d}{dx}\tan^{-1}(u) = \frac{1}{1+u^2}\frac{du}{dx}. \] Here, \( u = 3x^2y \). Its derivative is \[ \frac{d}{dx}(3x^2y) = 6xy + 3x^2y', \] so the derivative of the left-hand side becomes \[ \frac{6xy+3x^2y'}{1+(3x^2y)^2} = \frac{6xy+3x^2y'}{1+9x^4y^2}. \] 2. Differentiate the right-hand side. The derivative of \( x \) is \( 1 \) and for \( 2xy^2 \) we use the product rule: \[ \frac{d}{dx}(2xy^2) = 2y^2 + 4xyy'. \] Thus, the derivative of the right-hand side is \[ 1 + 2y^2 + 4xyy'. \] 3. Set the derivatives equal: \[ \frac{6xy+3x^2y'}{1+9x^4y^2} = 1 + 2y^2 + 4xyy'. \] 4. Multiply both sides by \( 1+9x^4y^2 \) to eliminate the denominator: \[ 6xy + 3x^2y' = \left(1+9x^4y^2\right)(1+2y^2 + 4xyy'). \] 5. Expand the right-hand side: \[ 6xy + 3x^2y' = \left(1+9x^4y^2\right)(1+2y^2) + 4xyy'\left(1+9x^4y^2\right). \] 6. Gather all terms with \( y' \) on one side: \[ 3x^2y' - 4xyy'\left(1+9x^4y^2\right) = \left(1+9x^4y^2\right)(1+2y^2) - 6xy. \] 7. Factor out \( y' \) on the left-hand side: \[ y'\left(3x^2 - 4xy\left(1+9x^4y^2\right)\right) = \left(1+9x^4y^2\right)(1+2y^2) - 6xy. \] 8. Finally, solve for \( y' \): \[ y' = \frac{\left(1+9x^4y^2\right)(1+2y^2)-6xy}{3x^2-4xy\left(1+9x^4y^2\right)}. \] The final answer is \[ y'=\frac{\left(1+9x^4y^2\right)(1+2y^2)-6xy}{3x^2-4xy\left(1+9x^4y^2\right)}. \]