Explanation

(c)
Let $y = {e^{x}}^{3}\mathcal{,z =}\log x$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = {e^{x}}^{3}\left( 3x^{2} \right) = 3x^{2}{e^{x}}^{3}$and $\frac{d\mathcal{z}}{dx} = \frac{1}{x}$
$\therefore\frac{dy}{d\mathcal{z}} = \frac{\frac{dy}{dx}}{\frac{d\mathcal{z}}{dx}} = \frac{3x^{2}{e^{x}}^{3}}{\left( \frac{1}{x} \right)} = 3x^{3}{e^{x}}^{3}$