Explanation

(b)
$\because x^{x}y^{y}z^{z} = c$
$\Rightarrow \ \ \ \ \ x\log{x + y\log{y + z\log{z = \log c}}}$
On differentiating partially w.r.t. $x\ ,\$ we get
$x.\frac{1}{x} + \log{x + z.\frac{1}{z}\ }\frac{\partial z}{\partial x} + \log{z\frac{\partial z}{\partial x} = 0}$
$\Rightarrow (1 + \log{z)\frac{\partial z}{\partial x} = - (1 + \log{x)}}$
$\Rightarrow \ \frac{\partial z}{\partial x} = - \left( \frac{1 + \log x}{1 + \log z} \right)$