Problem with Implicit Differenation Problem:
Find $\frac{d^{2}y}{dx^2}$ in terms of $x$ and $y$.
\begin{equation*}
y^2 = x^3
\end{equation*}
Answer:
\begin{eqnarray*}
2y \frac{dy}{dx} &=& 3x^2 \\
\frac{dy}{dx} &=& \frac{3x^2}{2y} \\
\frac{d^2y}{dx^2} &=& \frac{2y(6x) - 3x^2(2)}{4y^2} \\
\frac{d^2y}{dx^2} &=& \frac{6xy - 3x^2}{2y^2} \\
\frac{d^2y}{dx^2} &=& \frac{6xy - 3x^2}{2x^3} \\
\end{eqnarray*}
However, the back of the book gets:
\begin{equation*}
\frac{d^{2}y}{dx^2} = \frac{3x}{4y}
\end{equation*}
I am hoping that somebody here can point out my mistake.

When you took the derivative of 2y, you forgot the $\frac{dy}{dx}$.
