Answer
(a) Using the Chain Rule we obtain
$\frac{{dy}}{{dx}} = - \frac{{4y}}{{3{y^2} + 4x}}$
(b) $g'\left( 1 \right) = 3$
Work Step by Step
(a) We have ${y^3} + 4xy = 16$.
Let $y = r\left( x \right)$ be a function of $x$ that solves the equation ${y^3} + 4xy = 16$.
Write $f\left( {x,y} \right) = {y^3} + 4xy = 16$. The partial derivatives of $f$ are
$\frac{{\partial f}}{{\partial x}} = 4y$, ${\ \ \ }$ $\frac{{\partial f}}{{\partial y}} = 3{y^2} + 4x$
Using the Chain Rule we get
$\frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dx}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}} = 0$
$\frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}} = 0$
$4y + \left( {3{y^2} + 4x} \right)\frac{{dy}}{{dx}} = 0$
Hence, $\frac{{dy}}{{dx}} = - \frac{{4y}}{{3{y^2} + 4x}}$.
(b) We have $g\left( x \right) = f\left( {x,r\left( x \right)} \right)$ and
${f_x}\left( {1,2} \right) = 8$, ${\ \ \ }$ ${f_y}\left( {1,2} \right) = 10$
Using the Chain Rule we calculate $g'\left( 1 \right)$:
$g'\left( x \right) = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dx}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}} = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}}$
$g'\left( x \right) = {f_x} + \left( { - \frac{{4y}}{{3{y^2} + 4x}}} \right){f_y}$
Since $y = r\left( x \right)$ and $r\left( 1 \right) = 2$, so at $\left( {x,y} \right) = \left( {1,2} \right)$, we obtain
$g'\left( 1 \right) = 8 - \frac{{4\cdot2}}{{3\cdot{2^2} + 4\cdot1}}\cdot10$
$g'\left( 1 \right) = 3$
