The equation of the normal to the curve \(y=x \log x\) parallel to \(2 x-2 y+3=0\) is

\(\begin{aligned}
& y=x \log x ...(i)\\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=1+\log x
\end{aligned}\)
Slope of the normal \(=\frac{-1}{\frac{\mathrm{~d} y}{\mathrm{~d} x}}=-\frac{1}{1+\log x}\)
Slope of the given line is 1.
Since the normal is parallel to the given line.
\(\begin{aligned}
\therefore \quad & -\frac{1}{1+\log x}=1 \\
& \Rightarrow \log x=-2 \\
& \Rightarrow x=\mathrm{e}^{-2} \\
& y=-2 \mathrm{e}^{-2}
\end{aligned}\)
...[From (i)]
\(\therefore \quad\) Equation of the normal at \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\) is
\(\begin{aligned}
& y+2 \mathrm{e}^{-2}=1\left(x-\mathrm{e}^{-2}\right) \\
& \Rightarrow y+2 \mathrm{e}^{-2}=x-\mathrm{e}^{-2} \\
& \Rightarrow x-y=3 \mathrm{e}^{-2}
\end{aligned}\)