If \(e^y+x y=e\) the ordered pair \(\left(\frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)\) at \(x=0\) is equal to

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 \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\).