Given, $c^{\prime}+x y=c$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& e^y \frac{d y}{d x}+x \frac{d y}{d x}+1 \cdot y=0 \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{-y}{\left(e^y+x\right)}
\end{aligned}
$$
Again, differentiating Eq. (ii) w.r.t. $x$, we get
$$
\begin{aligned}
& e^y \frac{d^2 y}{d x^2}+e^v\left(\frac{d y}{d x}\right)^2+x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=0 \\
\Rightarrow \quad & \left(e^y+x\right) \frac{d^2 y}{d x^2}+e^y\left(\frac{d y}{d x}\right)^2+2 \frac{d y}{d x}=0
\end{aligned}
$$
Now, on putting $x=0$ in Eq. (i), we get
$$
e^y+0 \cdot y=e \Rightarrow e^y=e^1 \Rightarrow y=1
$$
On putting $x=0, y=1$ in Eq. (iii), we get
$$
\frac{d y}{d x}=\frac{-1}{e+0}=-\frac{1}{e}
$$
Now, putting $x=0, y=1$ and $\frac{d y}{d x}=\frac{-1}{e}$ in
Eq. (iv), we get
$$
\begin{aligned}
& \left(e^1+0\right) \frac{d^2 y}{d x^2}+e^{\prime}\left(-\frac{1}{e}\right)^2+2\left(-\frac{1}{e}\right)=0 \\
\Rightarrow \quad & e \frac{d^2 y}{d x^2}+\frac{1}{e}-\frac{2}{e}=0 \Rightarrow \frac{d^2 y}{d x^2}=\frac{1}{e^2}
\end{aligned}
$$
Hence, $\left(\frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)$ at $x=0$ is $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$.