,

\begin{eqnarray*}
\frac{d}{dx}\left(y^{3}+e^{2-x}+5x\right) & = & \frac{d}{dx}\left(3\right)\\
3y^{2}\frac{dy}{dx}+e^{2-x}\frac{d}{dx}\left(2-x\right)+5 & = & 0\\
3y^{2}\frac{dy}{dx}-e^{2-x}+5 & = & 0
\end{eqnarray*}

,

\begin{eqnarray*}
3y^{2}\frac{dy}{dx} & = & e^{2-x}-5\quad\cdots\left(1\right)\\
\frac{dy}{dx} & = & \frac{e^{2-x}-5}{3y^{2}}
\end{eqnarray*}

,

take $\frac{d}{dx}$ กับสมการ $(1)$

\begin{eqnarray*}
\frac{d}{dx}\left(3y^{2}\cdot\frac{dy}{dx}\right) & = & \frac{d}{dx}\left(e^{2-x}-5\right)\\
3y^{2}\cdot\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\cdot\frac{d}{dx}\left(3y^{2}\right) & = & e^{2-x}\frac{d}{dx}\left(2-x\right)\\
3y^{2}\cdot\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\cdot6y & = & -e^{2-x}
\end{eqnarray*}

,

\begin{eqnarray*}
3y^{2}\cdot\frac{d^{2}y}{dx^{2}} & = & -e^{2-x}-\frac{dy}{dx}\cdot6y\\
\frac{d^{2}y}{dx^{2}} & = & -\frac{e^{2-x}+\frac{dy}{dx}\cdot6y}{3y^{2}}
\end{eqnarray*}

,

\begin{eqnarray*}
y^{3}+e^{0}+10 & = & 3\\
y^{3} & = & -8\\
y & = & -2
\end{eqnarray*}

,

แทนค่า $x=2$ และ $y=-2$

$$\left.\frac{dy}{dx}\right|_{x=2}=\frac{e^{0}-5}{12}=-\frac{1}{3}$$

,

แทนค่า $x=2$, $y=-2$ และ $\left.\frac{dy}{dx}\right|_{x=2}=-\frac13$

\begin{eqnarray*}
\left.\frac{d^{2}y}{dx^{2}}\right|_{x=2} & = & -\frac{e^{0}+\left(-\frac{1}{3}\right)\cdot6\left(-2\right)}{12}\\
& = & -\frac{5}{12}
\end{eqnarray*}