The general solution of $x\dfrac{dy}{dx} = y$ How to find the general solution to this differential equation?
Since $x$ and $y$ can be zero, I think I need to consider four separate cases: $(x=0,y=0)$, $(x\neq 0,y=0)$, and so on.
But I don't know how to proceed: I found when $x=0$, $y=0$, what does this mean??
If I divide both side by $xy$, then $|y| = C|x|$, where both $x$ and $y$ are non zero values. But how to take care of the situation when either of $x$ or $y$ is zero?
 A: So the case $x=0$ just implies that $y(0)=0$. This is more of an additional boundary condition that has to taken into account.
The case $y=0$ is more interesting because it provides an additional solution (namely $y\equiv0$). Well it is not really an additional solution because it can be also achieved by generalising the constant $C=e^{C^*}$ from your computation.
Now the question is how much regularity you impose on your solution. If you expect your solution to be $C^1$ your only option is $y(x)=Cx$ for $C\in\mathbb{R}$. If you relax your assumptions on the regularity to e.g. lipschitz you can use the interesting point $y=0$ to glue together different branches of your solution and still fulfill the ODE in a pointwise sense.
$$
y(x)=\begin{cases}C_1x&x<0\\C_2x&x\geq0\end{cases}
$$
for $C_1,C_2\in\mathbb{R}$.
A: $x=0$ implies $y=0$, and $\dfrac{dy}{dx}$ is free.
$y=0$ implies either $x=0$ (already seen), or $\dfrac{dy}{dx}=0$, which is perforce true, and $x$ is free. Hence $y=0$ is a solution.
Now we can assume $x\ne0,y\ne0$ and write
$$\frac{dy}y=\frac{dx}x.$$
Taking the antiderivative of both members,
$$\log|y|+c_y=\log|x|+c_x$$ where $c_x,c_y$ are two arbitrary constant, which can be combined in one.
Then taking the antilogarithm,
$$|y|=c|x|,$$ or $$y=\pm c|x|$$ which indeed satisfies the original equation, provided we change the sign at $x=0$ to ensure continuity.
Finally, $$y=cx$$ with no restrictions on $x,y,c$.
A: 
But I don't know how to proceed: I found when $x=0$, $y=0$, what does this mean??

When you put $x=0$ in your ODE of form $by'(x)+ay(x)=c$ you are not looking for a solution of ODE rather you look for a condition like $ay(0)+by'(0)=c$. In your case, it gives $y(0)=0$.
Now for the reverse case of plugging $y=0$, you can observe that ODE gives $0=0$ which shows that $y(x)=0$ is a solution (trivial) satisfying the condition $y(0)=0$.

How to find the general solution to this differential equation?

The general solution can be simply found as follows:
$xy'-y=0\implies \frac{xy'-y}{x^2}=0, x\ne 0$
$\implies d(y/x)=c
\implies y=cx, c\in\mathbb R$
is the general solution satisfying $y(0)=0$
