\( \begin{array}{lll}\text { 14. Find } y^{\prime}: & \text { (a) } 2 x y^{2}-\sin (x y)=x^{2}+y^{3} & \text { (b) } y=\tan (x / y)\end{array} \)

To find \( y' \) for the given equations, we'll need to use implicit differentiation and some algebraic manipulation. For part (a), the equation is \( 2xy^2 - \sin(xy) = x^2 + y^3 \). Start by differentiating both sides with respect to \( x \): \[ \frac{d}{dx}(2xy^2) - \frac{d}{dx}(\sin(xy)) = \frac{d}{dx}(x^2) + \frac{d}{dx}(y^3). \] When differentiating \( 2xy^2 \), use the product rule: \[ 2(y^2 + 2xy\frac{dy}{dx}). \] For \( \sin(xy) \), apply the chain rule: \[ \cos(xy)(y + x\frac{dy}{dx}). \] Now, plugging in all the derivatives and simplifying gives you an equation in terms of \( y' \). For part (b), the equation \( y = \tan\left(\frac{x}{y}\right) \) leads to: \[ \frac{dy}{dx} = \sec^2\left(\frac{x}{y}\right)\left(\frac{1}{y}\frac{dy}{dx} - \frac{x}{y^2}\frac{dy}{dx} + \frac{1}{y}\right). \] Rearranging and solving for \( y' \) gives the required derivative expression. Now you can tackle finding \( y' \) for both parts – happy differentiating!