The exercise is to solve
$$x u'(x) = \delta(x).$$
By using the definitions
$$\begin{cases}(u'|\varphi) = (-u|\varphi') \\ (fu|\varphi) = (u|f\varphi) \end{cases}$$
we get to solve
$$(-u|\varphi + x\varphi') = \varphi(0).$$
How do one proceed? Obviously we can see that $(-\delta|\varphi)$ is a solution, but what about other distributions?
Setting $v=u'$ we get the equation $xv=\delta$. The general solution to this is $v=-\delta'+a\delta,$ where $a$ is a constant. To get the general solution to $xu' = \delta$ we therefore need to solve $u' = -\delta'+a\delta,$ which is solved by $u=-\delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=\delta$ is $u=-\delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-\delta'+a\delta$ to $xv=\delta$ can be seen as a particular solution $v=-\delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=c\delta.$