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

\(y=x \log x\)
Differentiating w.r.to \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=1+\log x\)
\(\therefore \quad\) Slope of tangent \(=1+\log x\)
Slope of given line \(=1\)
\(\therefore \quad\) Line of normal is parallel to given line
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=-1 \\
& \Rightarrow 1+\log x=-1 \\
& \Rightarrow \log x=-2 \\
& \Rightarrow x=\mathrm{e}^{-2}
\end{aligned}\)
Substituting \(x=\mathrm{e}^{-2}\) in given equation,
\(\begin{aligned}
y & =x \log x \\
& =\mathrm{e}^{-2} \log \mathrm{e}^{-2} \\
& =-2 \mathrm{e}^{-2}
\end{aligned}\)
\(\therefore \quad\) Equation of line passing through \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\) is
\(\begin{aligned}
& y+2 \mathrm{e}^{-2}=1\left(x-\mathrm{e}^{-2}\right) \\
& \Rightarrow x-y=3 \mathrm{e}^{-2}
\end{aligned}\)