Question I : "If we are given any random differential equation, how do we identify that it would have singular solution as well ?"
Answer: For this you have to check the uniqueness of solution (click here for the theorem) of the given differential solution. If there is unique solution then, the given differential equation does not have any singular solution.
If your given differential equation is of Clairaut's form , i.e., of the form $$y=px+f(p)\qquad \text{where $p=\frac{dy}{dx}$}$$ then you make a conclusion that this differential equation has singular solution (for example click here).
Question II : "If I somehow manage to solve by Linear DE form, I won't obtain the singular solutions."
Answer: Yes, linear differential equations does not have any singular solutions. For explanation click here.
Question III : "$~~\cdots$ give me examples of DE where multiple singular curves are obtained ?"
Answer: Yes, there may exists more than one singular solutions. For counter example click here.
$$y^2\left(y - x\frac{dy}{dx}\right) = x^4\left(\frac{dy}{dx}
\right)^2$$
$$y(x)=-4x^2\quad\text{is a singular solution.}$$
$$y(x)=0\quad\text{is a singular solution.}$$
$x(y)=0$ is not solution of the ODE because $\frac{dy}{dx}$ is infinite which doesn't agree with the ODE.
$(x=0, y=0)$ isn't a function but a point. Thus it isn't a solution of the ODE.
The general solution of the ODE is :
$$y(x)=\frac{c^2 x}{x-c}$$
The above singular solutions are the envelope of the general solution.
$$ $$
IN ADDITION, ANSWER TO THE COMMENT from Hari Prasad (Too long answer to be edited in the comments section) :
The argument raised is :
When $x=0$ , then $dx/dy =0$ , which means $1/p=0$ , then $y^2(y/p^2−x/p)−x^4= 0$ when $x=0$ (implicitly $1/p=0$) . So, even $x=0$ is satisfying the differential equation. Is this thinking wrong ?
Let me remind this well-known argument : Given an ODE $$f(x,y,p)=0\tag 1$$ and multiplying by $g(x,y,p)$ gives a different ODE : $$g(x,y,p)f(x,y,p)=0\tag 2$$ which is not equivalent to Eq.$(1)$.The solutions of $(2)$ are the solutions of $(1)$ and the solutions of $(3)$ : $$g(x,y,p)=0\tag 3$$
Thus a solution of $(2)$ is not necessarily a solution of $(1)$. $$ $$
In your argument the equation $$f(x,y,p)=y^2(y-xp)-x^4p^2=0\tag 4$$ is multiplied by $g(x,y,p)=\frac{1}{p^2}$. So the solutions of $$y^2(\frac{y}{p^2}-\frac{x}{p})-x^4=0\tag 5$$ are not only the solutions of $(4)$ but in addition the solutions of $$\frac{1}{p^2}=\left(\frac{dx}{dy}\right)^2=0\tag 6$$ which are $x(y)=c$ and among them $x(y)=0$.
The solution $x=0$ is solution of $(5)$ but not necessarily solution of $(4)$.
Thus your argument is not correct: It doesn't prove that $x(y)=0$ is solution of Eq.$(4)$.
Best Answer
Your equation has one clear singular solution which is $y(x) = 0$. It may also have other singular solutions, but if it does these are not so easy to find. You are correct to use the $p$-discriminant method, but in general solving via this method doesn't give singular solutions, only the locus of points in the curve $f(x,y,p) = 0$ which have constant $p$ (in your case, $f(x,y,p) := p^2-3xp+y^2$). So you shouldn't expect the $p-$discriminant that you found to solve the equation.
Based on my understanding from here, if you want to construct all singular solutions it's necessary to find both the $p$- and $c$-discriminants of your equation. This would require finding the general solution of your equation $\phi(x,y,c)$ in terms of integration constant $c$ to be able to find the $c$-discriminant. I spent a while trying to solve your equation and I couldn't find a general solution, and I also put it into Mathematica which doesn't know how to solve it either. I was surprised Mathematica didn't know, I guess the equation itself looks deceptively simple.
Here is my calculation of the $p-$discriminant, based on the analysis here. First $$\frac{\partial f}{\partial p} = 2p - 3x=0.$$ Then $$2 f - p \frac{\partial f}{\partial p} = 2 y^2 - 3 x p = 0\qquad\implies\qquad p = \frac{2 y^2}{3x}.$$Now substitute back into $f(x,y,p) = 0$ to find that $$\psi(x,y):=y^2 \left(\frac{4 y^2}{9 x^2}-1\right) = 0.$$ I believe that this $\psi(x,y)$ is the $p$-discriminant. I think importantly, this method avoids dividing by $y$ in coming to the solution, and we see that $y=0$ turns out to be a singular solution here, so we don't want to divide by $y$ while solving the system to arrive at the form for $\psi(x,y)$.
Then setting $\psi(x,y)$ to zero gives two solutions to the $p$-discriminant equation; the first one is $y(x) = 0$, and the second is $y(x)^2=\frac{9}{4}x^2$. This gives $|y(x)| = \pm \frac{3}{2} |x|$, which is almost what you have. But as I said, I think it's fine that this doesn't solve $f(x,y,y') = 0$.
Out of interest, where does this equation come from? Is it from some research problem, or from a course on differential equations?