Using the information in the tool box , let us substitute $y = vx$ and $dy/dx = v + xdv/dx.$
$v + x.dv/dx = (x + vx)/(x-vx)$
Taking $v$ as tha common factor in the RHS and cancelling we get
$v + x.dv/dx = (1+v)/(1-v)$
Bringing $v$ from the LHS to RHS we get,
$x.dv/dx = (1+v^2)/(1-v)$
Step3:
Separating the variables we get
$(1-v)dv/(1+v^2) = dx/x$
$dv/1+v^2 - vdv/1+v^2 = dx/x$
Integrating on both sides we get,
$tan^{-1}v - (1/2)log(1+v^2) = logx + C$
Step4:
Substituting back for v as y/x we get
$tan^{-1}(y/x) - (1/2)log(x^2+y^2)/x^2 = logx + C$
This can be rewritten as ,
$tan^{-1}(y/x) - (1/2)x^2+y^2 + (1/2)log x^2 = logx + C$
$tan^{-1}(y/x) = (1/2)log(x^2+y^2) + C$
This is the required solution.