To begin with, we clarify that by expression "p" we mean the first derivative with respect to the variable "x" :
"p=\\frac{dy}{dx}"
Now we rewrite the initial equation, expressing the variable "y" as a function of "x" and "p"
"x^3p^2+x^2yp+a^3=0\\longrightarrow\\\\[0.3cm]\n\\left. x^2yp=-a^3-x^3p^2\\right|\\div\\left(x^2p\\right)\\longrightarrow\\\\[0.3cm]\ny=-\\frac{a^3}{x^2p}-\\frac{x^3p^2}{x^2p}\\longrightarrow\\\\[0.3cm]\\boxed{y=-\\frac{a^3}{x^2p}-xp}"
We differentiate the resulting expression with respect to "x" , remembering that "p=dy\/dx" in order to exclude the variable "y" from the equation.
"\\frac{d}{dx}\\left|y=-\\frac{a^3}{x^2p}-xp\\right.\\longrightarrow\\\\[0.3cm]\np=-p-x\\cdot\\frac{dp}{dx}-\\frac{a^3}{x^3p}\\cdot(-2)-\\frac{a^3}{x^2p^2}\\cdot\\left(-\\frac{dp}{dx}\\right)\\\\[0.3cm]\np+p-\\frac{2a^3}{x^3p}=-\\left(x-\\frac{a^3}{x^2p^2}\\right)\\cdot\\frac{dp}{dx}\\\\[0.3cm]\n2p\\cdot\\left(1-\\frac{a^3}{x^3p^2}\\right)=-x\\cdot\\left(1-\\frac{a^3}{x^3p^2}\\right)\\cdot\\frac{dp}{dx}\\\\[0.3cm]\n2p=-x\\cdot\\displaystyle\\frac{dp}{dx}"
"2p=-x\\cdot\\frac{dp}{dx}\\longrightarrow\\frac{dp}{p}=-\\frac{dx}{2x}\\longrightarrow\\\\[0.3cm]\n\\ln|p|=-2\\cdot\\ln|x|+\\ln|C|\\longrightarrow\\\\[0.3cm]\n\\ln|p|+\\ln\\left|x^2\\right|=\\ln|C|\\longrightarrow px^2=C\\\\[0.3cm]\n\\boxed{p=\\frac{C}{x^2}}"
Substitute the resulting ratio in the original equation
"x^3p^2+x^2yp+a^3=0\\longrightarrow\\\\[0.3cm]\nx^3\\cdot\\left(\\frac{C}{x^2}\\right)^2+x^2y\\cdot\\left(\\frac{C}{x^2}\\right)+a^3=0\\longrightarrow\\\\[0.3cm]\n\\left.\\frac{C^2}{x}+Cy+a^3=0\\right|\\times(x)\\\\[0.3cm]\n\\boxed{C^2+(xy)C+a^3x=0}"
which is a quadratic equation in "C" . Equating it's discriminant to 0, we get :
"D=x^2y^2-4a^3x=0\\longrightarrow\\\\[0.3cm]\nx\\cdot\\left(xy^2-4a^3\\right)=0\\longrightarrow\n\\left[\\begin{array}{l}\nx=0\\\\[0.3cm]\nxy^2-4a^3=0\n\\end{array}\\right."
ANSWER
"\\left[\\begin{array}{l}\nx=0\\\\[0.3cm]\nxy^2-4a^3=0\n\\end{array}\\right."
are the singular solutions of the given differential equation.
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