Answer
$$y(x)=x^{x}+C x^{x} e^{-x}$$
Work Step by Step
Given $$y^{\prime}-(\ln x) y=x^{x}$$
This a linear equation with $p(x) = -\ln x$ and $q(x) =x^x$; then
\begin{align*}
\mu(x) &= e^{\int p(x)dx}\\
&=e^{\int -\ln xdx}\\
&=e^{-[x\ln x-x]}\\
&=e^{\ln x^{-x} +x}\\
&=x^{-x}e^x
\end{align*}
Hence, the general solution is
\begin{align*}
\mu(x) y &=\int \mu (x)xq(x)dx\\
x^{-x}e^xy&=\int x^{-x}e^x x^{x}dx\\
&=\int x^{-x}e^x x^{x}dx\\
&=\int e^{x}dx\\
&=e^{x}+C
\end{align*}
Hence $$y(x)=x^{x}+C x^{x} e^{-x}
$$
