Answer
\[y(x)=x\sin^{-1}(Cx)\]
Work Step by Step
\[x\frac{dy}{dx}=x\tan\left(\frac{y}{x}\right)+y\]
\[\frac{dy}{dx}=\tan\left(\frac{y}{x}\right)+\frac{y}{x}\;\;\;\ldots (1)\]
Substitute $\:y=Vx\;\;\;\ldots (2)$
Differentiate (2) with respect to $x$
\[\frac{dy}{dx}=V+x\frac{dV}{dx}\;\;\;\ldots (3)\]
From (1),(2) and (3)
\[V+x\frac{dV}{dx}=\tan V+V\]
\[x\frac{dV}{dx}=\tan V\]
Separating variables,
\[\cot V dV=\frac{1}{x}dx\]
Integrating,
\[\int\cot V dV=\int\frac{1}{x}dx+\ln C\]
Where $\ln C$ is constant of integration
\[\ln |\sin V|=\ln |x|+\ln |C|=\ln |Cx|\]
\[\sin V=Cx\]
\[V=\sin^{-1}(Cx)\]
From (2)
\[\frac{y}{x}=\sin^{-1}(Cx)\]
\[y=x\sin^{-1}(Cx)\]
Hence general solution of (1) is \[y(x)=x\sin^{-1}(Cx)\;.\]
