(g) $y = \\sin^{-2}(x^{3})$

Explanation:

Step1: Identify the outer - inner functions

Let $u = x^{3}$, then $y=\sin^{-2}(u)$. First, find the derivative of $y$ with respect to $u$ and $u$ with respect to $x$.

Step2: Differentiate $y$ with respect to $u$

Using the power - rule and the chain - rule for $y = (\sin(u))^{-2}$. Let $t=\sin(u)$. Then $y = t^{-2}$. The derivative of $y$ with respect to $t$ is $\frac{dy}{dt}=-2t^{-3}=-2(\sin(u))^{-3}$. The derivative of $t$ with respect to $u$ is $\frac{dt}{du}=\cos(u)$. By the chain - rule $\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}=-2(\sin(u))^{-3}\cos(u)$.

Step3: Differentiate $u$ with respect to $x$

Since $u = x^{3}$, then $\frac{du}{dx}=3x^{2}$.

Step4: Use the chain - rule to find $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=-2(\sin(u))^{-3}\cos(u)$ and $\frac{du}{dx}=3x^{2}$ into the formula, and replace $u$ with $x^{3}$. $\frac{dy}{dx}=-2(\sin(x^{3}))^{-3}\cos(x^{3})\cdot3x^{2}=-6x^{2}\frac{\cos(x^{3})}{\sin^{3}(x^{3})}$.

Answer:

$-6x^{2}\frac{\cos(x^{3})}{\sin^{3}(x^{3})}$