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Let us introduce the variable
$t={1\over x}$:
$$x^2\, y_{x}-y^2=x\, \Longrightarrow\ y_{t}+y^2+{1\over t}=0\ .$$
This is the so-called Riccati equation.
It is equivalent to a linear ODE. Indeed, if
$$\Psi=e^{\int^t dt\, y}\, , $$
then
$$\Psi_{tt}+{\Psi\over t}=0\ .$$
The last can be transformed to the Bessel equation:
$$\Psi=z\, F(z)\, ,\ \ z=2\sqrt{t}\ \ \Longrightarrow\
F_{zz}+{F_{z}\over z}+\Big(\, 1-{1\over z^2}\,\Big)=1\ .$$
Hence
$$F(z)=C_1\ J_1(z)+C_2\ Y_1(z)\ ,$$
where
$J_1(z),\ Y_1(z)$ are conventional Bessel functions.
($C_{1,2}$ are integration constants).
Now one has,
$$\Psi(t)=const\, \sqrt{t}\
\big(\, J_1(2\sqrt{t})+ c\ Y(2\sqrt{t})\, \big)\ .$$
Here $c$ is a constant.
Finally,
$$y={\Psi_t\over \Psi}=\sqrt{x}\ \ \
{J_0({2\over \sqrt{x}})+c\,Y_0({2\over \sqrt{x}})\over
J_1({2\over \sqrt{x}})+c\, Y_1({2\over \sqrt{x}}) }\ .$$
Here I use the identities,
$$J_1'(z)=J_0(z)-{J_1(z)\over z}\, ,\ \ \ \
Y_1'(z)=Y_0(z)-{Y_1(z)\over z}\ .$$
All the best.
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