Answer
(a) $y' =\frac{x}{y^2}$
(b) $y''=\frac{y^{3}-2 x^{2}}{y^{5}}$
Work Step by Step
We are given:
$$y^{3}-\frac{3}{2} x^{2}=1$$
(a) Differentiate both sides
\begin{align*}
3 y^{2} y^{\prime}-3 x&=0\\
y'&=\frac{x}{y^2}
\end{align*}
and
\begin{align*}
y''&= \frac{y^2-2xyy'}{y^4}
\end{align*}
(b) Since $y'=x/y^2$, then
\begin{align*}
y''&= \frac{y^2-2xyy'}{y^4}\\
&= \frac{y^2-2xy(x/y^2)}{y^4}\\
&= \frac{y^{3}-2 x^{2}}{y^{5}}
\end{align*}
